Compiler-Design

Question 1
Consider the grammar given below:
S → Aa
A → BD
B → b | ε
D → d | ε
Let a, b, d, and $ be indexed as follows:

Compute the FOLLOW set of the non-terminal B and write the index values for the symbols in the FOLLOW set in the descending order. (For example, if the FOLLOW set is {a, b, d, $}, then the answer should be 3210)

A
30
B
31
C
10
D
21
       Compiler-Design       GATE 2019
Question 1 Explanation: 
Follow(B) = First(D) Union Follow(A)
{Follow(B) = Follow(A) when D is epsilon}
Follow(B) = {d} Union {a} = {a,d}
Hence Answer is : 31
Question 2
Which one of the following kinds of derivation is used by LR parsers?
A
Leftmost in reverse
B
Rightmost in reverse
C
Leftmost
D
Rightmost
       Compiler-Design       GATE 2019
Question 2 Explanation: 
LR parsers have Rightmost derivation in reverse.
Question 3
Consider the augmented grammar given below:
    S' → S
    S → 〈L〉 | id
    L → L,S | S

Let I0 = CLOSURE ({[S' → ·S]}). The number of items in the set GOTO (I0 , 〈 ) is: _____.

A
4
B
5
C
6
D
7
       Compiler-Design       GATE 2019
Question 3 Explanation: 
I0 = CLOSURE ({[S' → ·S]})

Hence, the set GOTO (I0 , 〈 ) has 5 items.
Question 4

Consider the following grammar and the semantic actions to support the inheriteatd type declaration attributes. Let X1, X2, X3, X4, X5 and X6 be the placeholders for the non-terminals D, T, L or L1 in the following table:

     

Which one of the following are the appropriate choices for X1, X2, X3 and X4?

A
X1 = L, X2 = L, X3 = L1, X4 = T
B
X1 = L, X2 = T, X3 = L1, X4 = L
C
X1 = T, X2 = L, X3 = L1, X4 = T
D
X1 = T, X2 = L, X3 = T, X4 = L1
       Compiler-Design       GATE 2019
Question 4 Explanation: 
Since The production
L → L1, id {X3.type = X4.type } , this production has L and L1, hence X3 and X4 cannot be T.
So option 1, 3 and 4 cannot be correct.
Hence, 2 is correct answer.
Question 5
Which one of the following statements is FALSE?
A
Context-free grammar can be used to specify both lexical and syntax rules.
B
Type checking is done before parsing.
C
High-level language programs can be translated to different Intermediate Representations.
D
Arguments to a function can be passed using the program stack.
       Compiler-Design       Compilers       Gate 2018
Question 5 Explanation: 
Type checking is done in semantic analysis phase after syntax analysis phase (i.e., after parsing).
Question 6

Consider the following parse tree for the expression a#b$c$d#e#f, involving two binary operators $ and #.

Which one of the following is correct for the given parse tree?

A
$ has higher precedence and is left associative; # is right associative
B
# has higher precedence and is left associative; $ is right associative
C
$ has higher precedence and is left associative; # is left associative
D
# has higher precedence and is right associative; $ is left associative
       Compiler-Design       Associativity-and-Precedence       Gate 2018
Question 6 Explanation: 
Since $ will be evaluated before # so $ has higher precedence and the left $ i.e., in b$c$d the left “$” (i.e., b$c) will be evaluated first so it is left associative, whereas # is right associative (as in d#e#f) , the right one (i.e., e#f) will be evaluated first.
Question 7

A lexical analyzer uses the following patterns to recognize three tokens T1, T2, and T3 over the alphabet {a,b,c}.

    T1: a?(b∣c)*a
    T2: b?(a∣c)*b
    T3: c?(b∣a)*c

Note that ‘x?’ means 0 or 1 occurrence of the symbol x. Note also that the analyzer outputs the token that matches the longest possible prefix.

If the string bbaacabc is processes by the analyzer, which one of the following is the sequence of tokens it outputs?

A
T1T2T3
B
T1T1T3
C
T2T1T3
D
T3T3
       Compiler-Design       Compilers       Gate 2018
Question 7 Explanation: 
a? means either 0 or 1 occurrence of “a”, so we can write T1, T2 and T3 as:
T1 : (b+c)*a + a(b+c)*a
T2 : (a+c)*b + b(a+c)*b
T3 : (b+a)*c + c(b+a)*c
Now the string is: bbaacabc
Please NOTE:
Token that matches the longest possible prefix
We can observe that the longest possible prefix in string is : bbaac which can be generated by T3.
After prefix we left with “abc” which is again generated by T3 (as longest possible prefix).
So, answer is T3T3.
Question 8

Consider the following intermediate program in three address code

           p = a - b
           q = p * c
           p = u * v
           q = p + q

Which of the following corresponds to a static single assignment form of the above code?

 
A
B
C
D
       Compiler-Design       Static-single-assignment       Gate 2017 set-01
Question 8 Explanation: 
Static Single Assignment is used for intermediate code in compiler design.
In Static Single Assignment form (SSA) each assignment to a variable should be specified with distinct names.
We use subscripts to distinguish each definition of variables.
In the given code segment, there are two assignments of the variable p
p = a-b
p = u*v
and two assignments of the variable q
q = p*c
q = p+q
So we use two variables p1, p2 for specifying distinct assignments of p and q1, q2 for each assignment of q.
Static Single Assignment form(SSA) of the given code segment is:
p1 = a-b
q1 = p1 * c
p2 = u * v
q2 = p2 + q1
Note:
As per options, given in GATE 2017 answer is B.
p3 = a - b
q4 = p3 * c
p4 = u * v
q5 = p4 + q4
Question 9

Consider the following grammar:

What is FOLLOW(Q)?

 
A
{R}
B
{w}
C
{w, y}
D
{w, $}
       Compiler-Design       First-and-Follow       Gate 2017 set-01
Question 9 Explanation: 
In the production: P → xQRS,
FOLLOW (Q) = FIRST (R)
FIRST (R) = {w, ϵ} >br> Since FIRST (R) = {ϵ}, so FOLLOW (Q) → {w} ∪ FIRST(S)
FIRST(S) = {y}
So, FOLLOW (Q) = {w, y}
Question 10

Consider the following grammar:

    stmt    →  if expr then expr else expr; stmt | ȯ
    expr    →  term relop term | term
    term    →  id | number
    id      →  a | bc
    number  → [0-9]

where relop is a relational operator (e.g., <, >, …), ȯ refers to the empty statement, and if, then, else are terminals.

Consider a program P following the above grammar containing ten if terminals. The number of control flow paths in P is ________. For example, the program

     if e1 then e2 else e3

has 2 control flow paths, e1 e2 and e1 e3.

 
A
1024
B
1025
C
1026
D
1027
       Compiler-Design       Parsers       Gate 2017 set-01
Question 10 Explanation: 
To get 10 'if' we need to use grammar to get,
if then else ; stmt
if then else ; if then else . stmt
:
:
:
(keep doing 10 times to get 10 'if')
We know that every if statement has 2 control flows as given in question. Hence,
We have 2 control flow choices for 1st 'if'
We have 2 control flow choices for 2nd 'if'
:
:
:
We have 2 control flow choices for 10th 'if'
Since all the choices are in one single structure or combination, so total choices are
2 × 2 × 2 × ........ 10 times = 210 = 1024
Question 11

Consider the expression (a-1)*(((b+c)/3)+d)). Let X be the minimum number of registers required by an optimal code generation (without any register spill) algorithm for a load/store architecture, in which (i) only load and store instructions can have memory operands and (ii) arithmetic instructions can have only register or immediate operands. The value of X is ___________.

A
2
B
3
C
4
D
5
       Compiler-Design       Register-Allocation       Gate 2017 set-01
Question 11 Explanation: 
(a-1)*(((b+c)/3)+d)
Question 12

Match the following according to input (from the left column) to the compiler phase (in the right column) that processes it:

A
P→(ii), Q→(iii), R→(iv), S→(i)
B
P→(ii), Q→(i), R→(iii), S→(iv)
C
P→(iii), Q→(iv), R→(i), S→(ii)
D
P→(i), Q→(iv), R→(ii), S→(iii)
       Compiler-Design       Compilers       GATE 2017(set-02)
Question 12 Explanation: 
Character stream is input to lexical analyzer which produces tokens as output. So Q → (iv).
Token stream is forwarded as input to Syntax analyzer which produces syntax tree as output. So, S → (ii).
Syntax tree is the input for the semantic analyzer, So P → (iii).
Intermediate representation is input for Code generator. So R → (i).
Question 13

Which of the following statements about parser is/are CORRECT?

    I. Canonical LR is more powerful than SLR.
    II. SLR is more powerful than LALR.
    III. SLR is more powerful than Canonical LR.
A
I only
B
II only
C
III only
D
II and III only
       Compiler-Design       Parsers       GATE 2017(set-02)
Question 13 Explanation: 
Canonical LR is more powerful than SLR as every grammar which can be parsed by SLR parser, can also be parsed by CLR parser.
The power in increasing order is:
LR(0) < SLR < LALR < CLR
Hence only I is true.
Question 14

Consider the following expression grammar G:

    E → E - T | T
    T → T + F | F
    F → (E) | id

Which of the following grammars is not left recursive, but is equivalent to G?

A
B
C
D
       Compiler-Design       Left-Recursive-Grammar       GATE 2017(set-02)
Question 14 Explanation: 
Consider the production (given below) which has left recursion.
S→Sα | β
The equivalent production (after removing left recursion) is
S→βS1
S1→αS1 | ϵ
Hence after removing left recursion from: E→ E-T | T, here α = -T and β = T
E→TE1
E1→ -TE1 | ϵ
After removing left recursion from: T→T+F | F, here α=+F and β=F
T→FT1
T1→ +FT1 | ϵ
Replace E1 = X and T1 = Y
We have,
E→TX
X→-TX | ϵ
T→FY
Y→+FY | ϵ
F→(E)| id
Question 15

Consider the following code segment.

x = u - t;
y = x * v;
x = y + w;
y = t - z;
y = x * y;

The minimum number of  variables required to convert the above code segment to static single assignment form is ________.

 
A
10
B
11
C
12
D
13
       Compiler-Design       Static-single-assignment       2016 set-01
Question 15 Explanation: 
In Static Single Assignment form (SSA) each assignment to a variable should be specified with distinct names.
Generally, subscripts are used to distinguish each definition of variables.
In the given code segment, there are two assignments of the variable x
x = u - t;
x = y + w;
and three assignments of the variable y.
y = x * v;
y = t - z;
y = x * y
Hence, two variables viz x1, x2 should be used for specifying distinct assignments of x
and for y it is named as y1, y2 and y3 for each assignment of y.
Hence, total number of variables is 10 (x1, x2, y1, y2, y3, t, u, v, w, z), and there are 5 temporary variables.
Static Single Assignment form (SSA) of the given code segment is:
x1 = u - t;
y1 = x1 * v;
x2 = y1 + w;
y2 = t - z;
y3 = x2 * y2;
Question 16
 

The attributes of three arithmetic operators in some programming language are given below.

Operator      Precedence     Associativity     Arity 
   +             High            Left          Binary
   −            Medium           Right         Binary
   ∗             Low             Left          Binary

The value of the expression 2 – 5 + 1 – 7 * 3 in this language is __________.

A
9
B
10
C
11
D
12
       Compiler-Design       Associativity-and-Precedence       2016 set-01
Question 16 Explanation: 
+ has highest precedence, so it will be evaluated first.
2 − 5 + 1 − 7 * 3 = 2 − (5 + 1) − 7 * 3 = 2 − 6 − 7 * 3
Now, − has more precedence than *, so sub will be evaluated before * and – has right associative so (6 − 7) will be evaluated first.
2 − 6 − 7 * 3 = (2 − (6 − 7)) * 3 = (2 – (−1)) * 3 = 3 * 3 = 9
Question 17
 

Consider the following Syntax Directed Translation Scheme (SDTS), with non-terminals {S, A} and terminals {a,b}.

  S → aA   { print 1 }
  S → a    { print 2 }
  A → Sb   { print 3 }

Using the above SDTS, the output printed by a bottom-up parser, for the input aab is:

 
A
1 3 2
B
2 2 3
C
2 3 1
D
syntax error
       Compiler-Design       Syntax-Directed-Translation       2016 set-01
Question 17 Explanation: 
By using bottom up parser, the output will be “2 3 1”
Question 18

Match the following:

(P) Lexical analysis              (i) Leftmost derivation
(Q) Top down parsing             (ii) Type checking
(R) Semantic analysis           (iii) Regular expressions
(S) Runtime environments         (iv) Activation records
A
P ↔ i, Q ↔ ii, R ↔ iv, S ↔ iii
B
P ↔ iii, Q ↔ i, R ↔ ii, S ↔ iv
C
P ↔ ii, Q ↔ iii, R ↔ i, S ↔ iv
D
P ↔ iv, Q ↔ i, R ↔ ii, S ↔ iii
       Compiler-Design       Compilers       GATE 2016 set-2
Question 18 Explanation: 
Regular expressions are used in lexical analysis.
Top down parsing has left most derivation of any string.
Type checking is done in semantic analysis.
Activation records are loaded into memory at runtime.
Question 19

Which one of the following grammars is free from left recursion?

A
B
C
D
       Compiler-Design       Left-Recursive-Grammar       GATE 2016 set-2
Question 19 Explanation: 
The grammar in option A has direct left recursion because of production (A→Aa).
The grammar in option C has indirect left recursion because of production, (S→Aa and A→Sc).
The grammar in option D also has indirect left recursion because of production, (A→Bd and B→Ae).
Question 20

Which one of the following is True at any valid state in shift-reduce parsing?

 
A
Viable prefixes appear only at the bottom of the stack and not inside
B
Viable prefixes appear only at the top of the stack and not inside
C
The stack contains only a set of viable prefixes
D
The stack never contains viable prefixes
       Compiler-Design       Parsers       GATE 2015 (Set-01)
Question 20 Explanation: 
A handle is actually on the top of the stack.
A viable prefixes is prefix of the handle and so can never extend to the right of handle, i.e., top of stack.
So set of viable prefixes is in stack.
Question 21
The least number of temporary variables required to create a three-address code in static single assignment form for the expression q + r/3 + s – t * 5 + u * v/w is _________.
A
8
B
9
C
10
D
11
       Compiler-Design       Static-single-assignment       GATE 2015 (Set-01)
Question 21 Explanation: 
We will need one temporary variable for storing the result of every binary operation as Static Single Assignment implies the variable cannot be repeated on left hand side of assignment.
The given expression:
q+r/3+s−t∗5+u∗v/w
t1=r/3;
t2=t∗5;
t3=u∗v;
t4=t3/w;
t5=q+t1;
t6=t5+s;
t7=t6−t2;
t8=t7+t4;
So in total we need 8 temporary variables. If it was not mentioned as static single assignment then answer would have been 3 because we can re-use the same temporary variable several times.
Question 22
Let an represent the number of bit strings of length n containing two consecutive 1s. What is the recurrence relation for an?
A
an-2+an-1+2n-2
B
an-2+2an-1+2n-2
C
2an-2+an-1+2n-2
D
2an-2+2an-1+2n-2
       Compiler-Design       Live-Variable       GATE 2015 (Set-01)
Question 22 Explanation: 
For string of length 1, there is '0' consecutive 1's.
So, a1 = 0
For string of length 2,
a2 = 1
Similarly, a3 = 3
a4 = 8
Only (A) will satisfy the above values.
Question 23
A
p, s, u
B
r, s, u
C
r, u
D
q, v
       Compiler-Design       Live-Variable       GATE 2015 (Set-01)
Question 23 Explanation: 
In compilers, live variable analysis is a classic data flow analysis to calculate the variables that are live at each point in the program.
A variable is live at some point if it holds a value that may be needed in the future, of equivalently if its value may be read before the next time the variable is written to.
→ '1' can be assigned by the p and s and there is no intermediate use of them before that.
→ And p and s are not to be live in the both 2 & 3.
→ And q also assigned to u not live in 2 & 3.
→ And v is live at 3 not at 2.
→ u is live at 3 and also at 2, if we consider a path of length 0 from 2-8.
Finally r, u is the correct one.
Question 24
In the context of abstract-syntax-tree (AST) and control-flow-graph (CFG), which one of the following is TRUE?
A
In both AST and CFG, let node, N2 be the successor of node N1. In the input program, the code corresponding to N2 is present after the code corresponding in N1.
B
For any input program, neither AST nor CFG will contain a cycle
C
The maximum number of successors of a node in an AST and a CFG depends on the input program
D
Each node is AST and CFG corresponds to at most one statement in the input program
       Compiler-Design       Syntax-tree-and-context-flow-graph       GATE 2015 -(Set-2)
Question 24 Explanation: 
Optional (A) is not true when CFG contains cycle
Option (B) is false as CFG can contain cycle
Option (D) is false as a single node can contain block of statements.
Question 25
   
A
P-2,Q-3,R-1,S-4
B
P-2,Q-1,R-4,S-3
C
P-2,Q-4,R-1,S-3
D
P-2,Q-3,R-4,S-1
       Compiler-Design       Compilers       GATE 2015 -(Set-2)
Question 25 Explanation: 
P) Lexical analysis is related with FA and Regular expressions.
Q) Expression can be evaluated with postfix traversals.
R) Register allocation can be done by graph colouring.
S) The parser constructs a production tree.
Hence, answer is ( C ).
Question 26
 
A
5 and 7
B
6 and 7
C
5 and 5
D
7 and 8
       Compiler-Design       Control-Flow-Graph       GATE 2015 -(Set-2)
Question 26 Explanation: 
Question 27
Among simple LR (SLR), canonical LR, and look-ahead LR (LALR), which of the following pairs identify the method that is very easy to implement and the method that is the most powerful, in that order?
A
SLR, LALR
B
Canonical LR, LALR
C
SLR, canonical LR
D
LALR, canonical LR
       Compiler-Design       Parsers       GATE 2015(Set-03)
Question 27 Explanation: 
SLR is very easy to implement and CLR is most powerful method.
Question 28
       
A
Only S1
B
Only S2
C
Both S1 and S2
D
Neither S1 nor S2
       Compiler-Design       Parsers       GATE 2015(Set-03)
Question 28 Explanation: 
For LL(1),
For first production,

So, there is 'c' common in both the first(s) in the production of S. So not LL(1).
For LR(1),

Since R-R conflict is present. So, not LR(1).
Hence, Option (D) is the correct answer.
Question 29
Which one of the following is FALSE?
A
A basic block is a sequence of instructions where control enters the sequence at the beginning and exits at the end.
B
Available expression analysis can be used for common subexpression elimination.
C
Live variable analysis can be used for dead code elimination.
D
x=4*5⇒x=20 is an example of common subexpression elimination.
       Compiler-Design       Code-Optimization       GATE 2014(Set-01)
Question 29 Explanation: 
x=4* 5 ⇒ x=20 is an example of common subexpression elimination is a false statement.
Common subexpression elimination (CSE) is a compiler optimization that searches for instances of identical expressions (i.e., they all evaluate to the same value), and analyzes whether it is worthwhile replacing them with a single variable holding the computed value.
For ex: Consider the following code:
m=y+z * p
n= y+z *k
The common subexpression is “y+z” we can be calculate it one time and replace in both expression
temp = y+z
m = temp*p
n = temp*k
Question 30
A
a shift-reduce conflict and a reduce-reduce conflict.
B
a shift-reduce conflict but not a reduce-reduce conflict.
C
a reduce-reduce conflict but not a shift-reduce conflict.
D
neither a shift-reduce nor a reduce-reduce conflict.
       Compiler-Design       Parsers       GATE 2014(Set-01)
Question 30 Explanation: 
The input symbol is “<” which is not in canonical set of item, so it is neither a shift-reduce nor a reduce-reduce conflict with reference to “<” symbol.
But if it would have asked about “>” then it will be a SR conflict.
Question 31
A
Neither L nor is recursively enumerable (r.e.).
B
One of L and is r.e. but not recursive; the other is not r.e.
C
Both L and are r.e. but not recursive.
D
Both L and are recursive.
       Compiler-Design       Closure-Property       GATE 2014(Set-01)
Question 31 Explanation: 
If both L and L’ are recursively enumerable, then L must be recursive. Hence, both L and L´ are recursively enumerable, but not recursive is not a viable possibility.
Question 32
A
I and II only
B
I and III only
C
II and III only
D
I, II, and III
       Compiler-Design       Regular-Expressions       GATE 2014(Set-01)
Question 32 Explanation: 
The DFA accepts the language “all strings ending with 1”.
So the regular expression corresponding to DFA is (0+1)*1.
Now, by using state elimination method,
So the DFA also has another equivalent regular expression: 0*1(1+00*1)*.
But the regular expression (0*1*1+11*0*1) is not equivalent to DFA, as the DFA also accept the string “11011” which cannot be generated by this regular expression.
Question 33
A
+ is left associative, while ∗ is right associative
B
+ is right associative, while ∗ is left associative
C
Both + and ∗ are right associative
D
Both + and ∗ are left associative
       Compiler-Design       Associativity-and-Precedence       Gate 2014 Set -02
Question 33 Explanation: 
In production: T ⟶ T * U, since T is left recursive, hence * is left associative.
P ⟶ Q + P, here P is right recursive, so + is right associative.
Question 34
Which one of the following is NOT performed during compilation?
A
Dynamic memory allocation
B
Type checking
C
Symbol table management
D
Inline expansion
       Compiler-Design       Run-Time-Environments       Gate 2014 Set -02
Question 34 Explanation: 
Dynamic memory allocation is not performed during compilation, it occurs at run time only. At the time of compilation, compiler only compiles the instructions for dynamic memory allocation, like calloc, malloc….
Question 35
A
X is declared as “int X[32][32][8]”.
B
X is declared as “int X[4][1024][32]”.
C
X is declared as “char X[4][32][8]”.
D
X is declared as “char X[32][16][2]”.
       Compiler-Design       Intermediate-Code       Gate 2014 Set -02
Question 35 Explanation: 
Consider a 3-D array: arr[i][j][k]
Arr[0][1][2], this 3-D array contains
1 two dimensional array as i value is zero ( if i =1 then it has 2, two D array ), & the two dimension array has 2 row and 3 column.
So, In a 3-D array, i represent number of 2D arrays, j represent number of rows and k represent number of columns.
Number of 2 D array (M)=1
Number of rows (R)=2
Number of columns (C)=3

As arrays are stored in row major order, so this 2 dimension array will be stored as:

Assume base address of Arr is 1000. The address of required position is calculated as:
Arr[i][j][k]= Arr+ [i*(R*C)+(j*C)+k]*4 // multiplication to 4 is due to int takes 4 Bytes.
Arr[0][1][1]= 1000+[0*(2*3)+(1*3)+1]*4
=1000+[ 0+3+1 ]*4
=1000+4*4
=1016
As you can see that in the given example of row order storing of array also has address of Arr[0][1][1] is 1016.
Now coming to the question:
X [ i ][ j ][ k ] is calculated by 3 address code X[t4]
X [ i ][ j ][ k ] = X [ t4 ] // by substituting in reverse
=X [ t3 + t2]
=X [ t1 + t0 + k*4]
=X [ t0 + t1 + k*4] // t0 and t1 swapped as swapping doesn't have any impact
=X [ i*1024 + j*32 + k*4]
=X [ i*256 + j*8 +k] *4 // taking 4 (common) outside
= X [ i*(32*8)+ (j*8) +k] *4
By comparing the above line with ....... Arr[i][j][k]= Arr+ [i*(R*C)+(j*C)+k]*4
We get R=32, C=8
It means the the declared array has 32 rows and 8 columns and since multiplication by 4 (common outside) so it was declared as INT.
But how many number of 2D arrays in this 3D array, we don't know.
Since option A is the only option with configuration: INT arr[ ] [ 32 ] [ 8]. So it is right option.
Question 36
A
6
B
7
C
8
D
9
       Compiler-Design       General       Gate 2014 Set -02
Question 36 Explanation: 

Question 37
One of the purposes of using intermediate code in compilers is to
A
make parsing and semantic analysis simpler.
B
improve error recovery and error reporting.
C
increase the chances of reusing the machine-independent code optimizer in other compilers.
D
improve the register allocation.
       Compiler-Design       Code-Generation-and-Code-Optimization       Gate 2014 Set -03
Question 37 Explanation: 
Intermediate code is generated after semantic analysis. The intermediate code is independent of machine. By converting source code to intermediate code a machine independent code optimizer may be written. Thus increase the chances of reusing the machine-independent code optimizer in other compilers.
Question 38
A
1 and 2 only
B
2 and 3 only
C
3 and 4 only
D
1 and 3 only
       Compiler-Design       General       Gate 2014 Set -03
Question 38 Explanation: 
The statement, static allocation of all data areas by a compiler makes it impossible to implement recursion is true, as recursion requires memory allocation at run time, so it requires dynamic allocation of memory. Hence, Dynamic allocation of activation records is essential to implement recursion is also a true statement.
Question 39
A
6
B
7
C
8
D
9
       Compiler-Design       Page-Replacement-Algorithm       Gate 2014 Set -03
Question 39 Explanation: 
6 page faults will occur using LRU.
Question 40
What is the maximum number of reduce moves that can be taken by a bottom-up parser for a grammar with no epsilon- and unit-production (i.e., of type A → є and A → a) to parse a string with n tokens?
A
n/2
B
n-1
C
2n-1
D
2n
       Compiler-Design       Parsing-and-Syntax-Directed-Graph       Gate 2013
Question 40 Explanation: 
Since it is given that the grammar cannot have:
1) epsilon production
2) production of the form A → a
Consider the grammar:
S → Sa | a
If we were to derive the string “aaa” whose length is 3 then the number of reduce moves that would have been required are shown below:
S→ Sa
→Saa
→aaa
This shows us that it has three reduce moves. The string length is 3 and the number of reduce moves is also 3. So presence of such kinds of production might give us the answer “n” for maximum number of reduce moves. But these productions are not allowed as per the question.
Also note that if a grammar does not have unit production then the maximum number of reduce moves can not exceed “n” where “n” denotes the length of the string.
3) No unit productions
Consider the grammar:
S→ A
A→ B
B→C
C→a
If we were to derive the string “a” whose length is 1 then the number of reduce moves that would have been required are shown below:
S→ A
A→ B
B→C
C→a
This shows us that it has four reduce moves. The string length is 1 and the number of reduce moves is 4. So presence of such kind of productions might give us the answer “n+1” or even more, for maximum number of reduce moves. But these productions are not allowed as per the question.
Now keeping in view the above points suppose we want to parse the string “abcd”. (n = 4) using bottom-up parsing where strings are parsed finding the rightmost derivation of a given string backwards. So here we are concentrating on deriving right most derivations only.
We can write the grammar which accepts this string which in accordance to the question, (i.e., with no epsilon- and unit-production (i.e., of type A → є and A → B) and no production of the form A→a) as follows:
S→aB
B→bC
C→cd
The Right Most Derivation for the above is:
S → aB (Reduction 3)
→ abC (Reduction 2)
→ abcd (Reduction 1)
We can see here the number of reductions present is 3.
We can get less number of reductions with some other grammar which also doesn’t produce unit or epsilon productions or production of the form A→a:
S→abA
A→ cd
The Right Most Derivation for the above is:
S → abA (Reduction 2)
→ abcd (Reduction 1)
Hence 2 reductions.
But we are interested in knowing the maximum number of reductions which comes from the 1st grammar. Hence total 3 reductions as maximum, which is (n – 1) as n = 4 here.
Question 41
 
A
1 only
B
2 only
C
1 and 4 only
D
1, 2, 3 and 4
       Compiler-Design       Parsing-and-Syntax-Directed-Graph       Gate 2013
Question 41 Explanation: 
The two sets in LR(1) items can be merged if they only differ with look aheads symbols, so statement 1 is false.
In the given LR(1) items there is not any reduce move, so after merging it will not have SR conflict and hence statement 2 and statement 3 are false.
Statement 4 is also wrong, because goto is carried on Non-Terminals symbols, not on terminal symbols, and c is a terminal symbol.
Hence all statements are false.
Question 42
A
B
C
D
       Compiler-Design       Run-Time-Environments       Gate 2012
Question 42 Explanation: 

Main → A1 → A2 → A21 → A1
Since, Activation records are created at procedure exit time.
A1 & A2 are defined under Main ( ). So A1 & A2 access links are pointed to main.
A21 is defined under A2, hence its access link will point to A2.
Question 43
 
A
FIRST(A) = {a,b,ε} = FIRST(B)
FOLLOW(A) = {a,b}
FOLLOW(B) = {a,b,$}
B
FIRST(A) = {a,b,$}
FIRST(B) = {a,b,ε}
FOLLOW(A) = {a,b}
FOLLOW(B) = {$}
C
FIRST(A) = {a,b,ε} = FIRST(B)
FOLLOW(A) = {a,b}
FOLLOW(B) = ∅
D
FIRST(A) = {a,b} = FIRST(B)
FOLLOW(A) = {a,b}
FOLLOW(B) = {a,b}
       Compiler-Design       Parsers       Gate 2012
Question 43 Explanation: 
FIRST (P): is the set of terminals that begin the strings derivable from non terminal P. If P derives epsilon then we include epsilon in FIRST(P).
FOLLOW(P): is the set of terminals that can appear immediately to the right of P in some sentential form.
FIRST(A) = FIRST (S)
FIRST (S) = FIRST (aAbB) and FIRST (bAaB) and FIRST (ϵ)
FIRST(S) = {a, b, ϵ}
FIRST (B) = FIRST (S) = {a, b, ϵ}= FIRST (A)
FOLLOW(A) = {b} // because of production S→a A b B
FOLLOW(A) = {a} // because of production S→ b A a B
So FOLLOW (A) = {a, b}
FOLLOW (B) = FOLLOW (S) // because of production S→ a A b B
FOLLOW (S) = FOLLOW (A) // because of production S → A
So FOLLOW (S) = {$, a, b}= FOLLOW(B)
Question 44
A
E1: S → aAbB,A → S
E2: S → bAaB,B→S
E3: B → S
B
E1: S → aAbB,S→ ε
E2: S → bAaB,S → ε
E3: S → ε
C
E1: S → aAbB,S → ε
E2: S → bAaB,S→ε
E3: B → S
D
E1: A → S,S →ε
E2: B → S,S → ε
E3: B →S
       Compiler-Design       Parsers       Gate 2012
Question 44 Explanation: 
The entries in E1, E2 and E3 is related to S and B, so we have to take only those production which have S and B in LHS.
S→ aAbB | bAaB | ε
The production S→ aAbB will go under column FIRST (aAbB) = a, so S→ aAbB will be in E1.
S→ bAaB will go under column FIRST(bAaB) = b, so S→ bAaB will be in E2.
S→ ε will go under FOLLOW (S) = FOLLOW(B) = {a, b, $ } , So S→ ε will go in E1, E2 and under column of $.
So E1 will have: S→ aAbB and S→ ε.
E2 will have S→ bAaB and S→ ε.
Now, B→ S will go under FIRST (S) = {a, b, ε}
Since FIRST(S) = ε so B→ S will go under FOLLOW (B) = {a, b, $}
So E3 will contain B→ S.
Question 45
The lexical analysis for a modern computer language such as Java needs the power of which one of the following machine models in a necessary and sufficient sense?
A
Finite state automata
B
Deterministic pushdown automata
C
Non-Deterministic pushdown automata
D
Turing machine
       Compiler-Design       Compilers       Gate 2011
Question 45 Explanation: 
Lexical Analysis is implemented by finite automata
Question 46
In a compiler, keywords of a language are recognized during
A
parsing of the program
B
the code generation
C
the lexical analysis of the program
D
dataflow analysis
       Compiler-Design       Compilers       Gate 2011
Question 46 Explanation: 
Any identifier is also a token so it is recognized in lexical Analysis
Question 47
Consider two binary operators ‘’ and ‘’ with the precedence of operator being lower than that of the operator . Operator is right associative while operator , is left associative. Which one of the following represents the parse tree for expression (73432)?  
A
B
C
D
       Compiler-Design       Parsers       Gate 2011
Question 47 Explanation: 
7 ↓ 3 ↑ 4 ↑ 3 ↓ 2
⇒ 7 ↓ (3 ↑ (4 ↑ 3)) ↓ 2
⇒ 7 ↓ (3 ↑ (4 ↑ 3))) ↓ 2 as ↓ is left associative
Question 48
 
A
2
B
9
C
5
D
3
       Compiler-Design       Register-Allocation       Gate 2011
Question 48 Explanation: 
Load R1, a ; R1 ← M[a]
Load R2, b ; R2 ← M[b]
Sub R1, R2 ; R1 ← R1 – R2
Load R2, c ; R2 ← M[c]
Load R3, d ; R3 ← M[d]
Add R2, R3 ; R2 ← R2 + R3
Load R3, e ; R3 ← M[e]
Sub R3, 3 ; R3 ← R3 – R2
Add R1, R3 ; R1 ← R1 + R3
Total 3 Registers are required minimum
Question 49
Which data structure in a compiler is used for managing information about variables and their attributes?
A
Abstract syntax tree
B
Symbol table
C
Semantic stack
D
Parse Table
       Compiler-Design       Compilers       2010
Question 49 Explanation: 
Symbol tables are data structures that are used by compilers to hold information about source-program constructs. The information is collected incrementally by the analysis phases of a compiler and used by the synthesis phases to generate the target code. Entries in the symbol table contain information about an identifier such as its character string (or lexeme) , its type, its position in storage, and any other relevant information.
Question 50
Which languages necessarily need heap allocation in the runtime environment?
A
Those that support recursion
B
Those that use dynamic scoping
C
Those that allow dynamic data structures
D
Those that use global variables
       Compiler-Design       Run-Time-Environments       2010
Question 50 Explanation: 
Dynamic memory is allocated on the heap by the system. So the languages which allow dynamic data structure require heap allocation at runtime.
Question 51
A
2
B
3
C
4
D
6
       Compiler-Design       Register-Allocation       2010
Question 51 Explanation: 
Here a, b, and c all have 3 different values so we need atleast 3 registers r1, r2 and r3.
Assume 'a' is mapped to r1, 'b' to r2 and 'c' to r3.
d = a + b, after this line if u notice 'a' is never present on right hand side, so we can map 'd' to r1.
e = c + d, after this line 'd' is never present on rhs, so we can map 'e' to r1.
at this time mapping is
r1 --- e
r2 --- b
r3 --- c
We have 3 registers for e, b and c.
f = c + e
b = c + e
These two are essentially doing same thing, after these two line 'b' and 'f' are same so we can skip computing 'f' or need not give any new register for 'f'. And wherever 'f' is present we can replace it with 'b', because neither of 'f' and 'b' are changing after these two lines, so value of these will be 'c+e' till the end of the program.
At second last line "d = 5 + e"
here 'd' is introduced, we can map it to any of the register r1 or r3, because after this line neither of 'e' or 'c' is required. Value of 'b' is required because we need to return 'd+f', and 'f' is essentially equal to 'b'
finally code becomes
r1 = 1
r2 = 10
r3 = 20
r1 = r1 + r2
r1 = r3 + r1
r2 = r3 + r1
r2 = r3 + r1
r1 = r2 + r2
r3 = 5 + r1
return r3 + r2
Therefore minimum 3 registers needed.
Question 52
The grammar S → aSa|bS|c is
A
LL(1) but not LR(1)
B
LR(1) but not LR(1)
C
Both LL(1) and LR(1)
D
Neither LL(1) nor LR(1)
       Compiler-Design       Parsers       2010
Question 52 Explanation: 
The LL(1) parsing table for the given grammar is:

As there is no conflict in LL(1) parsing table, hence the given grammar is LL(1) and since every LL(1) is LR(1) also, so the given grammar is LL(1) as well as LR(1).
Question 53
     
A
P-4, Q-1, R-2, S-3
B
P-3, Q-1, R-4, S-2
C
P-3, Q-4, R-1, S-2
D
P-2, Q-1, R-4, S-3
       Compiler-Design       Compilers       2009
Question 53 Explanation: 
Lexical analysis phase uses regular expression to identify the pattern of tokens, PDA is used in CFG and hence syntax analysis phase is related to PDA. Data flow analysis is done in code optimization phase and register allocation is related to code generation phase.
Question 54
 
A
I and II
B
I and IV
C
III and IV
D
I, III and IV
       Compiler-Design       Parsers       2009
Question 54 Explanation: 
Statement II is false, as a programming language which allows recursion requires dynamic storage allocation. Statement III is false, as L-attributed definition (assume for instance the L-attributed definition has synthesized attribute only) can be evaluated in bottom up framework.
Statement I is true, as the bottom up and top down parser take O(n) time to parse the string , i.e. only one scan of input is required.
Statement IV is true,Code improving transformations can be performed at both source language and intermediate code level. For example implicit type casting is also a kind of code improvement which is done during semantic analysis phase and intermediate code optimization is a topic itself which uses various techniques to improve the code such as loop unrolling, loop invariant.
Question 55
Which of the following describes a handle (as applicable to LR-parsing) appropriately?
A
It is the position in a sentential form where the next shift or reduce operation will occur.
B
It is non-terminal whose production will be used for reduction in the next step.
C
It is a production that may be used for reduction in a future step along with a position in the sentential form where the next shift or reduce operation will occur.
D
It is the production p that will be used for reduction in the next step along with a position in the sentential form where the right hand side of the production may be found.
       Compiler-Design       Parsers       Gate-2008
Question 55 Explanation: 
A handle is the production p that will be used for reduction in the next step along with a position in the sentential form where the right hand side of the production may be found.
Question 56
Some code optimizations are carried out on the intermediate code because
A
They enhance the portability of the compiler to other target processors
B
Program analysis is more accurate on intermediate code than on machine code
C
The information from dataflow analysis cannot otherwise be used for optimization
D
The information from the front end cannot otherwise be used for optimization
       Compiler-Design       Compilers       Gate-2008
Question 56 Explanation: 
The code-optimization on intermediate code generation will always enhance the portability of the compiler to target processors. The main reason behind this is, as the intermediate code is independent of the target processor on which the code will be executed, so the compiler is able to optimize the intermediate code more conveniently without bothering the underlying architecture of target processor.
Question 57
 
A
II and V only
B
I, III and IV only
C
I, II and V only
D
II, III and V only
       Compiler-Design       Run-Time-Environments       Gate-2008
Question 57 Explanation: 
II. Multilevel access link (or display) arrangement is needed to arrange Activation Records only if the programming language being implemented has nesting of procedures and functions.
V. PL’s which permits a function to return a function as its result cannot be implemented with a stack-based storage allocation scheme for activation records.
II & V are True.
Question 58
An LALR(1) parser for a grammar G can have shift-reduce (S-R) conflicts if and only if
A
the SLR(1) parser for G has S-R conflicts
B
the LR(1) parser for G has S-R conflicts
C
the LR(0) parser for G has S-R conflicts
D
the LALR(1) parser for G has reduce-reduce conflicts
       Compiler-Design       Parsres       Gate-2008
Question 58 Explanation: 
LALR(1) parser is obtained by minimising the LR(1) parser and hence they both uses LR(1) items. Now consider if LR(1) parser has SR conflict, for ex:
Consider a state in LR(1) parser:
S-> x.yA, a
A-> x. , y
This has both shift and reduce conflict on symbol “y”.
Since LR(1) already has SR conflict , so resulting LALR(1) (after merging) will also have SR conflict.
Now if LR(1) doesn’t have SR conflict then it is guaranteed that the LALR(1) will never have SR conflict. The reason behind this is, as we merge those state only which has same set of canonical items except look ahead and the LR(1) canonical items has DFA (means from one state to other state the transition is from unique symbol) , so after merging also we will never see any shift conflict, only reduce-reduce may occur.
Hence An LALR(1) parser for a grammar G can have shift-reduce (S-R) conflicts if and only if the LR(1) parser for G has S-R conflicts.
Question 59
Which one of the following is a top-down parser?
A
Recursive descent parser.
B
Operator precedence parser.
C
An LR(k) parser.
D
An LALR(k) parser.
       Compiler-Design       Parsers       Gate-2007
Question 59 Explanation: 
Recursive descent parser is top down parser, while others are bottom up parser.
Question 60
 
A
it is left recursive
B
it is right recursive
C
it is ambiguous
D
it is not context-free
       Compiler-Design       Parsers       Gate-2007
Question 60 Explanation: 
The given grammar is not left recursive and also it is context free (Type 2 grammar), so option A and D is wrong. Being a right recursive grammar is not an issue for LL(1) grammar. So even if given grammar is right recursive, this is not a reason for NOT LL(1).
This grammar has two parse tree for string “ibt ibt aea”.
Question 61
 
A
Both P and Q are true
B
P is true and Q is false
C
P is false and Q is true
D
Both P and Q are false
       Compiler-Design       Parsers       Gate-2007
Question 61 Explanation: 
Every regular grammar is LL(1) is false, as the grammar may have left recursion or left factoring or also it is possible that grammar is ambiguous.
For ex: Consider a regular grammar
S->aS | a | ϵ
this grammar is ambiguous as for string "a" two parse tree is possible.

Hence it is regular but not LL(1).
But every regular set has a language acceptor as DFA , so every regular set must have atleast one grammar which is unambiguous.
Hence, every regular set has LR(1) grammar.
Question 62
 
A
aaaabb
B
aabbbb
C
aabbab
D
abbbba
       Compiler-Design       Grammar       Gate-2007
Question 62 Explanation: 
The string “aabbab” can be derived by following steps:
S -> aB [Using S --> aB]
-> aaBB [Using B --> aBB]
-> aabB [Using B --> b]
-> aabbS [Using B --> bS]
-> aabbaB [Using S --> aB]
-> aabbab [Using B --> b]
Question 63
A
1
B
2
C
3
D
4
       Compiler-Design       Grammar       Gate-2007
Question 63 Explanation: 
There exist two parse tree for the string “aabbab” using LMD (left most derivation)
Question 64
Consider an ambiguous grammar G and its disambiguated version D. Let the language recognized by the two grammars be denoted by L(G) and L(D) respectively. Which one of the following is true ?
A
L (D) ⊂ L (G)
B
L (D) ⊃ L (G)
C
L (D) = L (G)
D
L (D) is empty
       Compiler-Design       Grammar       Gate 2007-IT
Question 64 Explanation: 
By changing the corresponding grammar, the language will not be changed.
For example, by converting NFA to DFA language will not be changed.
Question 65
   
A
(i) and (ii)
B
(ii) and (iii)
C
(i) and (iii)
D
None of the above
       Compiler-Design       Parsers       Gate-2006
Question 65 Explanation: 
As we can see in the below given LR(0) items, that all three belongs to different state (sets).
Question 66
 
A
2 * 3 + 4
B
2 * +3 4
C
2 3 * 4 +
D
2 3 4+*
       Compiler-Design       Syntax-Directed-Translation       Gate-2006
Question 66 Explanation: 

Now perform post order evaluation, you will get output as,
2 3 4 + *
Question 67
 
A
The code contains loop invariant computation
B
There is scope of common sub-expression elimination in this code
C
There is scope of strength reduction in this code
D
There is scope of dead code elimination in this code
       Compiler-Design       Code-Optimization       Gate-2006
Question 67 Explanation: 
→ 4*j is a common sub-expression. So there is a scope of elimination. So B is correct.
→ 5*i can be moved out of inner loop. So can be i%2, here loopinvarient computation can be done, so option A is correct.
→ 4*i, 5*j can also be replaced so there is a scope of strength reduction. So C is true.
→ But there is no dead code to eliminate, we can replace the variable representation only.
Question 68
 
A
{S → FR} and {R → ε}
B
{S → FR} and { }
C
{S → FR} and {R → *S}
D
{F → id} and {R → ε}
       Compiler-Design       Parsers       Gate-2006
Question 68 Explanation: 
Predictive parsing table for the mentioned grammar:

The representation M[X,Y] means X represents Variable (rows) and Y represents terminals (columns).
The productions are filled in parsing table by the below mentioned rules:
For every production P → α, we have:
Rule 1: If P → α is a production then add this production for each terminal “t” which is in FIRST of [α] i.e., ADD P → α to M[P, a]
Rule 2: If “ϵ” belongs to FIRST of [P] then add P → α to M[P, b] where “b” represents terminals FOLLOW[P].
By the above rules, we can see that production S → FR will go M[S, a] where “a” is FIRST [FR] which is equal to FIRST[F] = id, So S → FR will go in M[S,id].
Since in the production R→ϵ , FIRST[ϵ] = ϵ, hence the production will go in M[R, b] where “b” represents terminals FOLLOW[R] and FOLLOW[R] = $, so production R→ϵ will go in M[R,$]
Question 69
 
A
S1 is false and S2 is false
B
S1 is false and S2 is true
C
S1 is true and S2 is false
D
S1 is true and S2 is true
       Compiler-Design       Code-Optimization       Gate-2006
Question 69 Explanation: 
We cannot compare the program on the basis of runtime with respect to any inputs.
So, given statement is wrong.
S1: Let us assume an array = {1,2,3,4,5} and i=0.
Let j = i+2 = 0+2 = 2
For the respective example work1 and work2 results 1 and 0, so S1 statement is False.
Question 70
The grammar A → AA | (A) | ε is not suitable for predictive-parsing because the grammar is:
A
ambiguous
B
left-recursive
C
right-recursive
D
an operator-grammar
       Compiler-Design       Parsers       Gate-2005
Question 70 Explanation: 
The given grammar can be turned into a infinite parse tree. So it is ambiguous.
It have A → AA has left recursion.
Question 71

A
It detects recursion and eliminates recursion
B
It detects reduce-reduce conflict, and resolves
C
It detects shift-reduce conflict, and resolves the conflict in favor of a shift over a reduce action
D
It detects shift-reduce conflict, and resolves the conflict in favor of a reduce over a shift action
       Compiler-Design       Parsres       Gate-2005
Question 71 Explanation: 
Yacc favours shift move in case of SR conflict.
Question 72
A
Equal precedence and left associativity; expression is evaluated to 7
B
Equal precedence and right associativity; expression is evaluated to 9
C
Precedence of '×' is higher than that of '+', and both operators are left associative; expression is evaluated to 7
D
Precedence of '+' is higher than that of '×', and both operators are left associative; expression is evaluated to 9
       Compiler-Design       Parsers       Gate-2005
Question 72 Explanation: 
First of all, it is ambiguous grammar. Hence, equal precedence and associativity. Now as Yacc resolved it with shift move we will shift until the last operator and then we will start reducing.

Hence, the answer is 9 and right associative.
Question 73
 
A
No compilation error
B
Only a lexical error
C
Only syntactic errors
D
Both lexical and syntactic errors
       Compiler-Design       Compilers       Gate-2005
Question 73 Explanation: 
There is no error in the above code. Actually it is a link error. Here compiler fro is a function which is not declared. Hence, it will not produce any error. It will only throw a warning in C.
Question 74
 
A
n, E + n and E + n × n
B
n, E + n and E + E × n
C
n, n + n and n + n × n
D
n, E + n and E × n
       Compiler-Design       Grammar       Gate-2005
Question 74 Explanation: 
E → E + n {Applying E → E+n}
→ E + E * n {Applying E → E * n}
→ E + n * n {Applying E → n}
→ n + n * n {Applying E → n}
We use n, E+n, E×n reductions to get a sentence n+n*n.
Question 75
   
A
n1 < n2 < n3
B
n1 = n3 < n2
C
n1 = n2 = n3
D
n1 ≥ n3 ≥ n2
       Compiler-Design       Parsers       Gate-2005
Question 75 Explanation: 
→ SLR(1) and LALR(1) both are be the states of LR(0) items then SLR(1) = LALR(1).
→ LR(1) be the states of LR(1) items.
→ LR(0) items never be greater than LR(1) items then SLR(1) = LALR(1) < LR(1)
n1 = (n3) < (n2)
Question 76
 
A
(i) only
B
(i) and (iii) only
C
(ii) and (iii) only
D
(iii) and (iv) only
       Compiler-Design       Parsers       Gate-2004
Question 76 Explanation: 
Operator values doesn't contains nullable values and two adjacent non-terminals on RHS production.
i) On RHS it contains two adjacent non-terminals.
ii) Have nullable values.
Question 77
Consider a program P that consists of two source modules M1 and M2 contained in two different files. If M1 contains a reference to a function defined in M2, the reference will be resolved at
A
Edit time
B
Compile time
C
Link time
D
Load time
       Compiler-Design       Compilers       Gate-2004
Question 77 Explanation: 
The link time can gives the reference to the executable file when the functions are present in the other modules.
Question 78
Consider the grammar rule E → E1 - E2 for arithmetic expressions. The code generated is targeted to a CPU having a single user register. The subtraction operation requires the first operand to be in the register. If E1 and E2 do not have any common sub-expression, in order to get the shortest possible code
A
E1 should be evaluated first
B
E2 should be evaluated first
C
Evaluation of E1 and E2 should necessarily be interleaved
D
Order to evaluation of E1 and E2 is of no consequence
       Compiler-Design       Target-Code-Generation       Gate-2004
Question 78 Explanation: 
After evaluating E2 first and then E1, we will have E2 in the register and then we can simply do SUB operation with E2 which will be in memory. And if we do E1 first and then E2, then we must move E2 to memory and again bring back E1 to the register before doing SUB operation, which will increase load.
Question 79
   
A
200
B
180
C
160
D
40
       Compiler-Design       Grammar       Gate-2004
Question 79 Explanation: 
Given expression is
2 # 3 & 5 # 6 & 4
→ Here # means multiplication (*)
& means addition (+)
→ & is having more precedence because it is far from starting symbol, here # and & are left associatives.
2 # 3 & 5 # 6 & 4
⇒ (2 * (3+5)) * (6+4)
⇒ (2 * 8) * (10)
⇒ 16 * 10 = 160
Question 80
     
A
{w|Na(w) > 3Nb(w)}
B
{w|Nb(w) > 3Na(w)}
C
{w|Na(w) = 3k, k Î {0, 1, 2, ...}}
D
{w|Nb(w) = 3k, k Î {0, 1, 2, ...}}
       Compiler-Design       Grammar       Gate-2004
Question 80 Explanation: 
S→bS
S→baA
S→babA
S→babaB
S→babaa
n(a)=3; n(b)=2
Option A:
Na(w) > 3Nb(w)
3 > 3(2)
3 > 6 (✖️)
Option B:
Nb(w) > 3Nb(w)
2 > 3(2)
2 > 6 (✖️)
Option D:
Nb(w) = 3k
2 = 3k(✖️)
S = aA
S→aA
S→abA
S→abaB
S→abaa
n(a)=3
|n(a)|=3
→ Answer: Option C(✔️)
Question 81
Which of the following suffices to convert an arbitrary CFG to an LL(1) grammar?  
A
Removing left recursion alone
B
Factoring the grammar alone
C
Removing left recursion and factoring the grammar
D
None of the above
       Compiler-Design       Grammar       Gate-2003
Question 81 Explanation: 
Left recursion removing (or) factoring the given grammar are not sufficient to convert an arbitrary CFG to an LL(1) grammar.
To convert an arbitrary CFG to an LL(1) grammar we need to remove the left recursion and as well as left factoring without that we cannot convert.
Question 82
Assume that the SLR parser for a grammar G has n1 states and the LALR parser for G has n2 states. The relationship between n1 and n2 is  
A
n1 is necessarily less than n2
B
n1 is necessarily equal to n2
C
n1 is necessarily greater than n2
D
None of the above
       Compiler-Design       Parsers       Gate-2003
Question 82 Explanation: 
No. of states in SLR and LALR are equal and no. of states in SLR and LALR are less than or equal to LR(1).
Question 83
In a bottom-up evaluation of a syntax directed definition, inherited attributes can  
A
always be evaluated
B
be evaluated only if the definition is L-attributed
C
be evaluated only if the definition has synthesized attributes
D
never be evaluated
       Compiler-Design       Syntax-Directed-Translation       Gate-2003
Question 83 Explanation: 
L-Attributed grammar can able to inherits either inherited attributes (or) synthesized attributes.
L-Attributed definitions are a class of syntax directed definitions whose attributes can be evaluated by a single traversal of the parse-tree.
Question 84
Which of the following statements is FALSE?  
A
In statically typed languages, each variable in a program has a fixed type
B
In un-typed languages, values do not have any types
C
In dynamically typed languages, variables have no types
D
In all statically typed languages, each variable in a program is associated with values of only a single type during the execution of the program
       Compiler-Design       Run-Time-Environments       Gate-2003
Question 84 Explanation: 
Dynamic typed languages are those languages in which variable must necessarily be defined before they are used. Then dynamic typed languages have types.
Question 85
 
A
{S'→e S} and {S'→ε}
B
{S'→e S} and { }
C
{S'→ε} and {S'→ε}
D
{S'→e S, S'→ε} and {S'→ε}
       Compiler-Design       Parsers       Gate-2003
Question 85 Explanation: 
First(S) = {1,a}
First(S') = {e,ε}
First(E) = {b}
Follow(S') = {e,$}
Only when 'First' contains ε, we need to consider FOLLOW for getting the parse table entry.

Hence, option (D) is correct.
Question 86
 
A
LL(1)
B
SLR(1) but not LL(1)
C
LALR(1) but not SLR(1)
D
LR(1) but not LALR(1)
       Compiler-Design       Parsers       Gate-2003
Question 86 Explanation: 

Hence, it is LL(1).
Question 87
 
A
9 + 5 + 2
B
9 5 + 2 +
C
9 5 2 + +
D
+ + 9 5 2
       Compiler-Design       Syntax-Directed-Translation       Gate-2003
Question 87 Explanation: 

Now traverse the tree and whatever comes first to print, just print it.
Answer will be 9 5 + 2 +.
Question 88

A
X = Y + Z
B
t1 = Y + Z; X = t1
C
t1= Y; t2 = t1 + Z; X = t2
D
t1 = Y; t2 = Z; t3 = t1 + t2; X = t3
       Compiler-Design       Syntax-Directed-Translation       Gate-2003
Question 88 Explanation: 
Question 89
Which of the following is NOT an advantage of using shared, dynamically linked libraries as opposed to using statically linked libraries?  
A
Smaller sizes of executable files
B
Lesser overall page fault rate in the system
C
Faster program startup
D
Existing programs need not be re-linked to take advantage of newer versions of libraries
       Compiler-Design       Run-Time-Environments       Gate-2003
Question 89 Explanation: 
Dynamic link libraries takes more time in program setup (in loading and linking phase to set up the global offset table and load and link the required libraries).
Question 90
 
A
Theory Explanation is given below.
       Compiler-Design       Descriptive       Gate-2002
Question 91
 
A
3
B
26
C
10
D
21
       Compiler-Design       Compilers       Gate-2000
Question 91 Explanation: 
We have six different types of tokens are available
(i) Keyword
(ii) Identifier
(iii) Constant
(iv) Variable
(v) String
(vi) Operator
Print = Token 1
( = Token 2
"i=%d%x" = Token 3 [Anything inside " " is one Token]
, = Token 4
i = Token 5
, = Token 6
& = Token 7
i = Token 8
) = Token 9
; = Token 10
Here, totally 10 Tokens are present in the equation.
Question 92
Which of the following derivations does a top-down parser use while parsing an input string? The input is assumed to be scanned in left to right order.
A
Leftmost derivation
B
Leftmost derivation traced out in reverse
C
Rightmost derivation
D
Rightmost derivation traced out in reverse
       Compiler-Design       Parsers       Gate-2000
Question 92 Explanation: 
Top-down parser - Leftmost derivation
Bottom-Up parser - Reverse of rightmost derivation
Question 93
 
A
* has higher precedence than +
B
- has higher precedence than *
C
+ and – have same precedence
D
+ has higher precedence than *
       Compiler-Design       Grammar       Gate-2000
Question 93 Explanation: 
The operator which is in low level that can have high preference.
Order of precedence is *, +, -.
Here * and + have equal preference, '-' can have higher precedence than + and *.
Question 94

A
Theory Explanation is given below.
       Compiler-Design       Descriptive       Gate-2000
Question 95
Which of the following is the most powerful parsing method?
A
LL (1)
B
Canonical LR
C
SLR
D
LALR
       Compiler-Design       Parsers       Gate-1999
Question 95 Explanation: 
Canonical LR is most powerful.
LR > LALR > SLR
Question 96
The number of tokens in the Fortran statement DO 10 I = 1.25 is
A
3
B
4
C
5
D
None of the above
       Compiler-Design       Compilers        Gate-1999
Question 96 Explanation: 
DO → 1
10 → 2
I → 3
= → 4
1.25 → 5
Question 97
In a resident – OS computer, which of the following systems must reside in the main memory under all situations?
A
Assembler
B
Linker
C
Loader
D
Compiler
       Compiler-Design       Run-Time-Environments       Gate-1998
Question 97 Explanation: 
In many operating system loader is permanently resident in memory.
Some OS may allow virtual memory may allow the loader to be located in a region of memory that is in page table.
Question 98
Which of the following statements is true?
A
SLR parser is more powerful than LALR
B
LALR parser is more powerful than Canonical LR parser
C
Canonical LR parser is more powerful than LALR parser
D
The parsers SLR, Canonical CR, and LALR have the same power
       Compiler-Design       Parsers       Gate-1998
Question 98 Explanation: 
LR > LALR > SLR
Canonical LR parser is more powerful than LALR parser.
Question 99
Type checking is normally done during
A
lexical analysis
B
syntax analysis
C
syntax directed translation
D
code optimization
       Compiler-Design       Compilers       Gate-1998
Question 99 Explanation: 
Type checking is normally done during syntax directed translation.
Question 100
 
A
‘⊕’ is left associative while ‘*’ is right associative
B
Both ‘⊕’ and ‘*’ is left associative
C
‘⊕’ is right associative while ‘*’ is left associative
D
None of the above
       Compiler-Design       Grammar       Gate-1997
Question 100 Explanation: 

⊕ is left associative.
* is right associative.
Question 101

A language L allows declaration of arrays whose sizes are not known during compilation. It is required to make efficient use of memory. Which one of the following is true?

A
A compiler using static memory allocation can be written for L
B
A compiler cannot be written for L; an interpreter must be used
C
A compiler using dynamic memory allocation can be written for L
D
None of the above
       Compiler-Design       Run-Time-Environments       Gate-1997
Question 101 Explanation: 
Compiler is use dynamic memory allocation then the memory will be allocated to an array at runtime.
Question 102
The conditional expansion facility of macro processor is provided to
A
test a condition during the execution of the expanded program
B
to expand certain model statements depending upon the value of a condition during the execution of the expanded program
C
to implement recursion
D
to expand certain model statements depending upon the value of a condition during the process of macro expansion
       Compiler-Design       Macros       Gate-1997
Question 102 Explanation: 
Macro is expanded during the process of Macro expansion.
Question 103
Heap allocation is required for languages
A
that support recursion
B
that support dynamic data structures
C
that use dynamic scope rules
D
None of the above
       Compiler-Design       Run-Time-Environments       Gate-1997
Question 103 Explanation: 
Heap allocation is required for languages that support dynamic data structures.
Question 104
 
A
1, 2, 1, 2
B
2, 1, 2, 1
C
2, 1, 1, 2
D
1, 2, 2, 2
       Compiler-Design       Assembler       Gate-1996
Question 104 Explanation: 
The functionalities from pass 1 and pass 2 are:
Pass 1:
1) Assign addresses to all statements in the program.
2) Save the values assigned to all labels for use in pass 2.
3) Perform some processing of assembler directives.
Pass 2:
1) Assemble instructions.
2) Generate data values defined by BYTE, WORD etc.
3) Perform processing of assembler directives not done during pass 1.
4) Write the program and assembling listing.
Question 105
 
A
(ii) only
B
(i) only
C
both (i) and (ii)
D
None of the above
       Compiler-Design       Macros       Gate-1996
Question 105 Explanation: 
If M2 macro is called with X=0, then it will go into an infinite loop.
Question 106
A linker is given object modules for a set of programs that were compiled separately. What information need to be included in an object module?
A
Object code
B
Relocation bits
C
Names and locations of all external symbols defined in the object module
D
Absolute addresses of internal symbols
       Compiler-Design       Run-Time-Environments       Gate-1995
Question 106 Explanation: 
In object module it includes names and locations of all external symbols defined in the object module.
To link to external symbols it must know the location of external symbols.
Question 107
   
A
23131
B
11233
C
11231
D
33211
       Compiler-Design       Parsers       Gate-1995
Question 107 Explanation: 

⇒ 23131
Note SR is bottom up parser.
Question 108
Generation of intermediate code based on an abstract machine model is useful in compilers because
A
it makes implementation of lexical analysis and syntax analysis easier
B
syntax-directed translations can be written for intermediate code generation
C
it enhances the portability of the front end of the compiler
D
it is not possible to generate code for real machines directly from high level language programs
       Compiler-Design       Compilers       Gate-1994
Question 108 Explanation: 
In Intermediate code optimizations can also enhances the probability of optimizer.
Question 109
 
A
(i) - (d), (ii) - (a), (iii) - (b), (iv) - (c)
       Compiler-Design       General       Gate-1994
Question 109 Explanation: 
Backus Normal Form (BNF) is a notation technique for context free grammars, often used to describe the syntax of languages used in computing.
Yacc (Yet Another Compiler- Compiler) is a computer program for the UNIXoperating system. It is a LALR parser generator, generating a parser, the part of a compiler that tries to make syntactic sense of the source code, specially a LALR parser, based on an analytic grammar. Yacc is written in portable C.
Question 110
 
A
The program leads to compile time error
B
The program leads to run time error
C
The program outputs 5.2
D
The program produces error relating to nil pointer dereferencing
E
None of the above
       Compiler-Design       Compilers       Gate-1993
Question 110 Explanation: 
Note: Out of syllabus.
Question 111
A part of the system software, which under all circumstances must reside in the main memory, is:
A
text editor
B
assembler
C
linker
D
loader
E
none of the above
       Compiler-Design       General       Gate-1993
Question 111 Explanation: 
In a program the loader that can loads the object of the program from secondary memory into the main memory to execute the corresponding program. Then the loader is to be resides in the main memory.
Question 112
Consider the SLR(1) and LALR (1) parsing tables for a context free grammar. Which of the following statements is/are true?
A
The go to part of both tables may be different.
B
The shift entries are identical in both the tables.
C
The reduce entries in the tables may be different.
D
The error entries in the tables may be different.
E
B, C and D.
       Compiler-Design       Parsers       Gate-1992
Question 112 Explanation: 
Goto parts and shift entry must be same.
Reduce entry and error entry may be different due to conflicts.
Question 113
The process of assigning load addresses to the various parts of the program and adjusting the code and date in the program to reflect the assigned addresses is called
A
Assembly
B
Parsing
C
Relocation
D
Symbol resolution
       Compiler-Design       Run-Time-Environments       Gate-2001
Question 113 Explanation: 
Relocation can change the assigned address of data and code in the program.
Question 114
Which of the following statements is false?
A
An unambiguous grammar has same leftmost and rightmost derivation
B
An LL(1) parser is a top-down parser
C
LALR is more powerful than SLR
D
An ambiguous grammar can never be LR(k) for any k
       Compiler-Design       Parsers       Gate-2001
Question 114 Explanation: 
Option B: LL parser is a top-down parser for a subset of context-free languages. It parses the input from Left to right, performing Left most derivation of the sentence.
Option C: LALR is more powerful than SLR.
Option D: An ambiguous grammar can never be LR (k) for any k, because LR(k) algorithm aren’t designed to handle ambiguous grammars. It would get stuck into undecidability problem, if employed upon an ambiguous grammar, no matter how large the constant k is.
Question 115
 
A
Theory Explanation is given below.
       Compiler-Design       Parsers       Gate-2001
Question 116

A
Theory Explanation is given below.
       Compiler-Design       Syntax-Directed-Translation       Gate-2001
Question 117
   
A
Theory Explanation is given below.
       Compiler-Design       Grammar       Gate-2001
Question 118
The arithmetic expression : (a + b) * c - d / e * * f is to be evaluated on a two-address machine, where each operands, the number of registers required to evaluate this expression is ______. The number of memory access of operand is __________.
A
3 , 4
       Compiler-Design       Operands       Gate-1991
Question 118 Explanation: 
** is used for exponentiation.
So, in total 3 registers are required and 6 memory operations in total to fetch all operands.
Question 119
A given set of processes can be implemented by using only parbegin/parend statement, if the precedence graph of these processes is ________
A
properly nested.
       Compiler-Design       Precedence-Graph       Gate-1991
Question 119 Explanation: 
A given set of processes canf th be implemented by using only parbegin/parends statement, if the presedence graph of these processes is properly nested.
Question 120
 
A
matches the parameters of the macro-definition with locations of the parameters of the macro call
B
matches external names of one program with their location in other programs
C
matches the parameters of subroutine definition with the location of parameters of subroutine call
D
acts as link between text editor and the user
E
acts as a link between compiler and user program.
       Compiler-Design       Linker-Loader       Gate-1991
Question 120 Explanation: 
Linked editor can be able to perform
1) external symbol resolution
2) relocation
Question 121
     
A
Recursive descent parsing cannot be used for grammar with left recursion.
B
The intermediate form the representing expressions which is best suited for code optimization is the post fix form.
C
A programming language not supporting either recursion or pointer type does not need the support of dynamic memory allocation.
D
Although C does not support call by name parameter passing, the effect can be correctly simulated in C.
E
No feature of Pascal violates strong typing in Pascal.
F
A and D
       Compiler-Design       Parsers       Gate-1991
Question 121 Explanation: 
(A) It is true. Left recursive grammar if used directly in recursive descent parsing causes an infinite loop. So, left recursion must be removed before giving to a recursive descent parser.
(B) False.
(C) It is false. The language can have dynamic data types which required dynamically growing memory when data type size increases.
(D) Is true and using macro we can do this.
(E) Out of syllabus now.
Question 122
 
A
(a) - (q), (b) - (r), (c) - (s), (d) - (p)
       Compiler-Design       Match-the-Following       Gate-1990
Question 122 Explanation: 
Pointer data type - Dynamic data structure
Activation record - Recursion
Repeat until - Non-deterministic loop
Coercion - Type conversion
Question 123
 
A
(a) - (s), (b) - (p), (c) - (q), (d) - (r)
       Compiler-Design       Match-the-Following       Gate-1990
Question 123 Explanation: 
Lexical analysis - Finite automaton
Code optimization - DAG
Code generation - Syntax tree
Abelian groups - Push down automaton
Question 124
Merging states with a common core may produce __________ conflicts and does not produce ___________ conflicts in an LALR purser.  
A
Reduce-Reduce, Shift-Reduce
       Compiler-Design       Parsers       Gate-1989
Question 124 Explanation: 
Merge states with a common core may produce Reduce-Reduce conflicts and does not produce Shift-Reduce conflicts in an LALR parser.
Question 125
In a compiler the module the checks every character of the source text is called:
A
The code generator.
B
The code optimiser.
C
The lexical analyser.
D
The syntax analyser.
       Compiler-Design       Compilers       GATE-1987
Question 125 Explanation: 
Lexical analyzer phase checks every character of text to identify tokens.
Question 126
An operator precedence parser is a
A
Bottom-up parser.
B
Top-down parser.
C
Back tracking parser.
D
None of the above.
       Compiler-Design       Parsers       GATE-1987
Question 126 Explanation: 
An operator precedence parser is a Bottom-up parser.
Question 127
Using longer identifiers in a program will necessarily lead to:
A
Somewhat slower compilation
B
A program that is easier to understand
C
An incorrect program
D
None of the above
       Compiler-Design       Compilers       GATE-1987
Question 127 Explanation: 
Lexical analyzer will take more time to recognize the longer identifiers.
There are 127 questions to complete.