Programming
Question 1 
The return value of fun2 (5) is _______.
55 
int fun1(int n) {
printf("fun1 call\n");
static int i = 0;
if(n>0){
++i;
printf("fun1(%d1)\n",n);
fun1(n1);
}
printf("fun1(%d)= %d\n",n, i);
return(i);
}
int fun2(int n) {
printf("\n******* fun2 call ********\n");
static int i = 0;
if(n>0){
printf("%d + fun1(%d)\n", i,n);
i=i+fun1(n);
fun2(n1);
}
printf("fun2(%d)= %d\n",n, i);
return(i);
}
void main()
{
printf("final = %d\n", fun2(5));
}
Check step by step handrun of the code to understand the recursion:
******* fun2 call ********
0 + fun1(5)
fun1 call
fun1(51)
fun1 call
fun1(41)
fun1 call
fun1(31)
fun1 call
fun1(21)
fun1 call
fun1(11)
fun1 call
fun1(0)= 5
fun1(1)= 5
fun1(2)= 5
fun1(3)= 5
fun1(4)= 5
fun1(5)= 5
******* fun2 call ********
5 + fun1(4)
fun1 call
fun1(41)
fun1 call
fun1(31)
fun1 call
fun1(21)
fun1 call
fun1(11)
fun1 call
fun1(0)= 9
fun1(1)= 9
fun1(2)= 9
fun1(3)= 9
fun1(4)= 9
******* fun2 call ********
14 + fun1(3)
fun1 call
fun1(31)
fun1 call
fun1(21)
fun1 call
fun1(11)
fun1 call
fun1(0)= 12
fun1(1)= 12
fun1(2)= 12
fun1(3)= 12
******* fun2 call ********
26 + fun1(2)
fun1 call
fun1(21)
fun1 call
fun1(11)
fun1 call
fun1(0)= 14
fun1(1)= 14
fun1(2)= 14
******* fun2 call ********
40 + fun1(1)
fun1 call
fun1(11)
fun1 call
fun1(0)= 15
fun1(1)= 15
******* fun2 call ********
fun2(0)= 55
fun2(1)= 55
fun2(2)= 55
fun2(3)= 55
fun2(4)= 55
fun2(5)= 55
final = 55
Question 2 
The value returned by pp (3, 4) is ________.
81 
a=3,b=4
tot=1
ex=a=3
len=tob(b,arr) which is 3
[
tob(4,arr)==>
b=4
b%2 =4%2=0 Condition is false then arr[0]=0
=> b=b/2 =4/2 =2
b=2
b%2 =2%2=0 condition is false then arr[1]=0
=>b=b/2=2/2=1
b=1
then b%2=1%2 condition is true then arr[2]=1
=>b=b/2=1/2=0
The i value is 3 [length is 3]
]
i=0,
arr[0] ==1 condition is false
ex=3*3=9
i=1
arr[1]==1 condition is false
then
ex=9*9=81
i=2
then arr[2]==1 condition is true
tot=tot*ex=1*81=81
ex=81*81
Finally it returns tot value which 81
Question 3 
Consider the following C program:
#include <stdio.h> int counter = 0; int calc (int a, int b) { int c; counter++; if (b==3) return (a*a*a) ; else { c = calc (a, b/3) ; return (c*c*c) ; } } int main () { calc (4, 81); printf ("%d", counter) ; }
The output of this program is ______.
4  
5  
6  
7 
Question 4 
Consider the following C program.
#include <stdio.h> struct Outnode { char x, y, z ; }; int main () { struct Ournode p = {'1', '0', 'a'+2} ; struct Ournode *q = &p ; printf ("%c, %c", *((char*)q+1), *((char*)q+2)) ; return 0 ; }
The output of this program is
0, c  
0, a+2  
‘0’, ‘a+2’  
‘0’, ‘c’ 
The x variable here stores a character ‘c’ in it.
Because +2 will increment ascii value of a from 92 to 95.
Hence the structure p contains 3 character values and they are ‘1’, ‘0’, and ‘c’.
q is a pointer pointing to structure p.
Hence q is pointing to ‘1’, q+1 pointing to ‘0’ and q+2 pointing to ‘c’.
Option d cannot be correct, as though they are characters, printf will not print them in single quotes.
Question 5 
Consider the following C program:
#include<stdio.h> void fun1(char *s1, char *s2) { char *tmp; tmp = s1; s1 = s2; s2 = tmp; } void fun2(char **s2, char **s2) { char *tmp; tmp = *s1; *s1 = *s2; *s2 = tmp; } int main () { char *str1 = "Hi", *str2 = "Bye"; fun1(str1, str2); printf("%s %s", str1, str2); fun2(&str1, &str2); printf("%s %s", str1, str2); return 0; }
The output of the program above is
Hi Bye Bye Hi  
Hi Bye Hi Bye  
Bye Hi Hi Bye  
Bye Hi Bye Hi 
Hence, any change in the formal parameters are NOT reflected in actual parameters.
Hence, str1 points at “hi” and str2 points at “bye”.
The second call to the function ‘func2(&str1, &str2);’ is call by reference.
Hence, any change in formal parameters are reflected in actual parameters.
Hence, str1 now points at “bye” and str2 points at “hi”.
Hence answer is “hi bye bye hi”.
Question 6 
Consider the following C code. Assume that unsigned long int type length is 64 bits.
unsigned long int fun(unsigned long int n) { unsigned long int i, j, j=0, sum = 0; for (i = n; i > 1; i = i/2) j++; for ( ; j > 1; j = j/2) sum++; return sum; }The value returned when we call fun with the input 2^{40} is
4  
5  
6  
40 
Next for loop will divide j value (which is 40) by 2, each time until j>1.
j loop starts:
j=40 & sum=1
j=20 & sum=2
j=10 & sum=3
j=5 & sum=4
j=2 & sum=5
j=1 & break
So, sum = 5.
Question 7 
Consider the following program written in pseudocode. Assume that x and y are integers.
Count (x, y) { if (y != 1) { if (x != 1) { print("*") ; Count (x/2, y) ; } else { y = y  1 ; Count (1024, y) ; } } }
The number of times that the print statement is executed by the call Count(1024, 1024) is ______.
10230  
10231  
10232  
10233 
int count=0;
Count(x,y){
if(y!=1){
if(x!=1){
printf("*");
count = count +1;
Count(x/2,y);
}
else{
y=y1;
Count(1024,y);
}
}
}
void main()
{
Count(1024,1024);
printf("\n%d\n",count);
}
Count ( ) is called recursively for every (y = 1023) & for every y, Count ( ) is called (x = 10) times = 1023 × 10 = 10230
Question 8 
Consider the following C code:
#include <stdio.h> int *assignval(int *x, int val) { *x = val; return x; } void main ( ) { int *x = malloc(sizeof(int)); if(NULL == x) return; x = assignval(x, 0); if(x) { x = (int *)malloc(size of(int)); if(NULL == x) return; x = assignval(x, 10); } printf("%dn", *x); free(x); }
The code suffers from which one of the following problems:
compiler error as the return of malloc is not typecast approximately  
compiler error because the comparison should be made as x==NULL and not as shown  
compiles successfully but execution may result in dangling pointer  
compiles successfully but execution may result in memory leak 
In C++, we need to perform type casting, but in C Implicit type casting is done automatically, so there is no compile time error, it prints10 as output.
Option B:
NULL means address 0, if (a == 0) or (0 == a) no problem, though we can neglect this, as it prints 10.
Option C:
x points to a valid memory location. Dangling Pointer means if it points to a memory location which is freed/ deleted.
int*ptr = (int*)malloc(sizeof(int));
free(ptr); //ptr becomes a dangling pointer
ptr = NULL; //Removing Dangling pointers condition
Option D:
x is assigned to some memory location
int*x = malloc(sizeof(int));
→ (int*)malloc(sizeof(int)) again assigns some other location to x, previous memory location is lost because no new reference to that location, resulting in Memory Leak.
Hence, Option D.
Question 9 
Consider the following two functions.
void fun1(int n) { void fun2(int n) { if(n == 0) return; if(n == 0) return; printf("%d", n); printf("%d", n); fun2(n  2); fun1(++n); printf("%d", n); printf("%d", n); } }
The output printed when fun1(5) is called is
53423122233445  
53423120112233  
53423122132435  
53423120213243 
In fun2, we increment (pre) the value of n, but in fun1, we are not modifying the value.
Hence increment in value in recursion (back).
Hence, 5 3 4 2 3 1 2 2 2 3 3 4 4 5.
Question 10 
Consider the C functions foo and bar given below:
int foo(int val) { int x = 0; while (val>0) { x = x + foo(val); } return val; } int bar(int val) { int x = 0; while (val>0) { x = x + bar(val1); } return val; }
Invocations of foo(3) and bar(3) will result in:
Return of 6 and 6 respectively.  
Infinite loop and abnormal termination respectively.  
Abnormal termination and infinite loop respectively.  
Both terminating abnormally. 
{
x = x + foo(val);
}
In this case foo(val) is same as foo(val) & val ;
Because the recursive function call is made without changing the passing argument and there is no Base condition which can stop it.
It goes on calling with the same value ‘val’ & the system will run out of memory and hits the segmentation fault or will be terminated abnormally.
The loop will not make any difference here.
while(val>0)
{
x = x + bar(val1);
}
bar(3) calls bar(2)
bar(2) calls bar(1)
bar(1) calls bar(0) ⇾ Here bar(0) will return 0.
bar(1) calls bar(0)
bar(1) calls bar(0)……..
This will continue.
Here is a problem of infinite loop but not abrupt termination.
Some compilers will forcefully preempt the execution.
Question 11 
#include <stdio.h> #include <string.h> void printlength (char*s, char*t) { unsigned int c = 0; int len = ((strlen(s)  strlen(t)) > c) ? strlen(s):strlen(t); printf("%d\n", len); } void main () { char*x = "abc"; char*y = "defgh"; printlength(x,y); }Recall that strlen is defined in string.h as returning a value of type size_t, which is an unsigned int. The output of the program is _________.
3  
4  
5  
6 
{
char*x = "abc";
char*y = "defgh";
printlength(x,y);
}
printlength(char*3, char*t)
{
unsigned int c = 0;
int len = ((strlen(s)  strlen(t))> c) ? strlen(s) : strlen(t);
printf("%d", len);
}
Here strlen(s)  strlen(t) = 3  5 = 2
But in C programming, when we do operations with two unsigned integers, result is also unsigned. (strlen returns size_t which is unsigned in most of the systems).
So this result '2' is treated as unsigned and its value is INT_MAX2.
Now the comparison is in between large number & another unsigned number c, which is 0.
So the comparison returns TRUE here.
Hence (strlen(s)  strlen(t))>0 will return TRUE, on executing, the conditional operator will return strlen(s)
⇒ strlen(abc) = 3
Question 12 
The output of executing the following C program is __________.
#include<stdio.h> int total (int v) { static int count=0; while(v) { count += v&1; v ≫= 1; } return count; } void main() { static int x = 0; int i = 5; for(; 1> 0; i) { x = x + total(i); } printf("%d\n", x); }
23  
24  
25  
26 
Question 13 
P→(ii), Q→(iv), R→(i), S→(iii)  
P→(ii), Q→(i), R→(iv), S→(iii)  
P→(ii), Q→(iv), R→(iii), S→(i)  
P→(iii), Q→(iv), R→(i), S→(ii) 
⇾ A variable located in Data Section of memory
P→(ii), Q→(iv), R→(i), S→(iii)
Question 14 
Consider the following function implemented in C:
void printxy (int x, int y) { int *ptr; x = 0; ptr = &x; y = *ptr; *ptr = 1; printf("%d,%d",x,y); }
The output of invoking printxy(1, 1) is
0, 0  
0, 1  
1, 0  
1, 1 
{
int *ptr;
x = 0;
ptr = &x;
y = *ptr;
*ptr = 1;
}
printxy (1, 1)
Question 15 
Consider the C program fragment below which is meant to divide x and y using repeated subtractions. The variables x, y, q and r are all unsigned int.
while (r >= y) { r = r  y; q = q + 1; }
Which of the following conditions on the variables x, y, q and r before the execution of the fragment will ensure that the loop terminates in a state satisfying the condition x == (y*q + r)?
(q == r) && (r == 0)  
(x > 0) && (r == x) && (y > 0)  
(q == 0) && (r == x) && (y > 0)  
(q == 0) && (y > 0) 
x, y, q, r are unsigned integers.
while (r >= y)
{
r = r – y;
q = q + 1;
}
Loop terminates in a state satisfying the condition
x == (y * q + r)
y ⇒ Dividend = Divisor * Quotient + Remainder
So, to divide a number with repeated subtractions, the Quotient should be initialized to 0 and it must be incremented for every subtraction.
So initially q=0 which represents
x = 0 + r ⇒ x = r
and y must be a positive value (>0).
Question 16 
Consider the following snippet of a C program. Assume that swap(&x, &y) exchanges the contents of x and y.
int main () { int array[] = {3, 5, 1, 4, 6, 2}; int done = 0; int i; while (done == 0) { done = 1; for (i=0; i<=4; i++) { if (array[i] < array[i+1]) { swap (&array[i], &array[i+1]); done = 0; } } for (i=5; i>=1; i) { if (array[i] > array[i1]) { swap(&array[i], &array[i1]); done=0; } } } printf("%d", array[3]); }
The output of the program is ___________.
3  
4  
5  
6 
(1) ⇒ 1st for i = 0 <= 4
a[0] < a[1] ≃ 3<5 so perform swapping
done =
(1) ⇒ no swap (3, 1)
(2) ⇾ perform swap (1, 4)
(1) ⇒ perform swap (1, 6)
(1) ⇒ perform swap (1, 2)
(1) ⇒ (done is still 0)
for i = 5 >= 1 a[5] > a[4] ≃ 1>2 – false. So, no swapping to be done
(2) ⇾ no swap (6, 2)
(3) ⇾ swap (4, 6)
(1) ⇒ Swap (3, 6)
(1) ⇒ Swap (5, 6)
⇒ Done is still 0. So while loop executes again.
(1) ⇾ no swap (6, 5)
done = 1
(2) ⇾ No swap (5, 3)
(3) ⇾Swap (3, 4)
So, array [3] = 3
Question 17 
#include int main () { int m=10; int n, n1; n=++m; n1=m++; n; n1; n=n1; printf(“%d”, n); return 0; }The output of the program is ______
0  
1  
2  
3 
Question 18 
Consider the following C program.
#include<stdio.h> #include<string.h> int main () { char* c = "GATECSIT2017"; char* p = c; printf("%d", (int) strlen (c + 2[p]  6[p]  1)); return 0; }
The output of the program is __________.
1  
2  
4  
6 
char * P = C;
(int) strlen (C + 2[P] – 6[P] – 1)
C + 2[P]  6[P] – 1
∵2[P] ≃ P[2]
100 + P[2] – P[6] – 1
100 + T – I – 1
100 + 84 – 73 – 1
ASCII values: T – 84, I – 73
100 + 11 – 1
= 110
(int) strlen (110)
strlen (17) ≃ 2
Question 19 
Consider the following C program.
void f(int, short); void main () { int i = 100; short s = 12; short *p = &s; __________ ; // call to f() }
Which one of the following expressions, when placed in the blank above, will NOT result in a type checking error?
f(s, *s)  
i = f(i, s)  
f(i, *s)  
f(i, *p) 
short s = 12;
short *p = &s;
_______ // call to f ( ) :: (void f(int,short);)
It is clearly mentioned the return type of f is void.
By doing option elimination
(A) & (C) can be eliminated as s is short variable and not a pointer variable.
(B) i = f(i, s) is false because f’s return type is void, but here shown as int.
(D) f(i, *p)
i = 100
*p = 12
Hence TRUE
Question 20 
Consider the following C program.
#include<stdio.h> void mystery(int *ptra, int *ptrb) { int *temp; temp = ptrb; ptrb = ptra; ptra = temp; } int main() { int a=2016, b=0, c=4, d=42; mystery(&a, &b); if (a < c) mystery(&c, &a); mystery(&a, &d); printf("%d\n", a); }
The output of the program is ________.
2016  
2017  
2018  
2019 
For the first mystery (&a, &b);
temp = ptr b
ptr b = ptr a
ptr a = temp
If (a
Hence, a = 2016 will be printed.
Question 21 
The following function computes the maximum value contained in an integer array p[] of size n (n >= 1).
int max(int *p, int n) { int a=0, b=n1; while (__________) { if (p[a] <= p[b]) {a = a+1;} else {b = b1;} } return p[a]; }
The missing loop condition is
a != n  
b != 0  
b > (a + 1)  
b != a 
{
int arr [ ] = {3, 2, 1, 5, 4};
int n = sizeof(arr) / sizeof (arr[0]);
printf (max(arr, 5));
}
int max (int *p, int n)
{
int a = 0, b = n – 1;
(while (a!=b))
{
if (p[a] <= p[b])
{
a = a + 1;
}
else
{
b =b – 1;
}
}
return p[a];
}
The function computes the maximum value contained in an integer array p [ ] of size n (n >= 1).
If a = = b, means both are at same location & comparison ends.
Question 22 
What will be the output of the following C program?
void count(int n) { static int d=1; printf("%d ", n); printf("%d ", d); d++; if(n>1) count(n1); printf("%d ", d); } void main() { count(3); }
3 1 2 2 1 3 4 4 4  
3 1 2 1 1 1 2 2 2  
3 1 2 2 1 3 4  
3 1 2 1 1 1 2 
Count (3)
static int d = 1
It prints 3, 1
d++; //d = 2
n>1, count(2)
prints 2, 2
d++; // d = 3
n>1, count(1)
prints 1, 3 → Here n = 1, so condition failed & printf (last statement) executes thrice & prints d
d++; //d=4 value as 4. For three function calls, static value retains.
∴ 312213444
Question 23 
What will be the output of the following pseudocode when parameters are passed by reference and dynamic scoping is assumed?
a=3; void n(x) {x = x * a; print(x);} void m(y) {a = 1; a = y  a; n(a); print(a);} void main() {m(a);}
6, 2  
6, 6  
4, 2  
4, 4 
First m(a) is implemented, as there are no local variables in main ( ), it takes global a = 3;
m(3) is passed to m(y).
a = 1
a = 3 – 1 = 2
n(2) is passed to n(x).
Since it is dynamic scoping
x = 2 * 2 = 4 (a takes the value of its calling function not the global one).
The local x is now replaced in m(y) also.
Hence, it prints 4,4.
And we know it prints 6, 2 if static scoping is used.
It is by default in C programming.
Question 24 
The value printed by the following program is __________.
void f(int* p, int m) { m = m + 5; *p = *p + m; return; } void main() { int i=5, j=10; f(&i, j); printf("%d", i+j); }
30  
31  
32  
33 
P is a pointer stores the address of i, & m is the formal parameter of j.
Now, m = m + 5;
*p = *p + m;
Hence, i + j will be 20 + 10 = 30.
Question 25 
The following function computes X^{Y} for positive integers X and Y.
int exp (int X, int Y) { int res = 1, a = X, b = Y; while ( b != 0 ){ if ( b%2 == 0) { a = a*a; b = b/2; } else { res = res*a; b = b1; } } return res; }
Which one of the following conditions is TRUE before every iteration of the loop?
X^{Y} = a^{b}  
(res * a)^{Y} = (res * X)^{b}  
X^{Y} = res * a^{b}  
X^{Y} = (res * a)^{b} 
{
int res = 1, a = X, b = Y;
while (b != 0)
{
if (b%2 == 0)
{
a = a*a;
b = b/2;
}
else
{
res = res*a;
b = b – 1;
}
}
return res;
}
From that explanation part you can understand the exponent operation, but to check the conditions, first while iteration is enough.
x = 2, y = 3, res = 2, a = 2, b = 2.
Only (C) satisfies these values.
x^{y} = res * a^{b}
2^{3} = 2 * 2^{2} = 8
Explanation:
Will compute for smaller values.
Let X = 2, Y = 3, res = 1
while (3 != 0)
{
if(3%2 == 0)  False
else
{
res = 1*2 = 2;
b = 3 – 1 = 2;
}
For options elimination, consider
return res = 2 (but it is out of while loop so repeat while)
__________
while (2 != 0)
{
if (2%2 == 0)  True
{
a = 2*2 = 4
b = 2/2 = 1
}
__________
repeat while
while (1 != 0)
{
if (1%2 == 0)  False
else
{
res = 2 * 4 = 8
b = 1 – 1 = 0
}
__________
while (0 != 0)  False
return res = 8 (2^{3})
Question 26 
Consider the following program:
int f(int *p, int n) { if (n <= 1) return 0; else return max (f(p+1,n1),p[0]p[1]); } int main() { int a[] = {3,5,2,6,4}; printf("%d", f(a,5)); }Note: max(x,y) returns the maximum of x and y.
The value printed by this program is __________.
3  
4  
5  
6 
f(a, 5) ⇒ f(100, 5)
Question 27 
void f1 (int a, int b) { int c; c=a; a=b; b=c; } void f2 (int *a, int *b) { int c; c=*a; *a=*b;*b=c; } int main() { int a=4, b=5, c=6; f1(a, b); f2(&b, &c); printf (“%d”, cab); return 0; }
5  
6  
7  
8 
But f_{2} will swap the value of 'b' and 'c' because f_{2} is call by reference. So finally the value of
a=4
b=6
c=5
So, answer will be
c  a  b
5  4  6 = 5
Question 28 
#include <stdio.h> int main() { unsigned int x[4][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}, {10, 11, 12}}; printf("%u, %u, %u", x+3, *(x+3), *(x+2)+3); }
2036, 2036, 2036  
2012, 4, 2204  
2036, 10, 10  
2012, 4, 6 
⇒ x [4] [3] can represents that x is a 2dimensional array.
⇒ x+3 = (Address of x) + 3 * 4 * 3 [3×4×3 is inner dimention]
= 2000 + 36
= 2036
⇒ *(x+3) also returns the address i.e., 2036.
The '*' represents 1  D but x is starting at 2036.
⇒ *(x+3)+3 = *(Address of x + 2 * 3 * 4) + 3
= *(2000 + 24) +3
= *(2024) + 3 ['*' will change from 2D to 1D]
= 2024 + 3 * 4
= 2024 + 12
= 2036
Question 29 
begin q := 0 r := x while r >= y do begin r := r – y q := q + 1 end endThe post condition that needs to be satisfied after the program terminates is
{r = qx+y ∧ r  
{x = qy+r ∧ r  
{y = qx+r ∧ 0  
{q+1 
⇒ x = qy + r
Question 30 
while (first <= last) { if (array [middle] < search) first = middle +1; else if (array [middle] == search) found = True; else last = middle – 1; middle = (first + last)/2; } if (first < last) not Present = True;The cyclomatic complexity of the program segment is __________.
5  
6  
7  
8 
Question 31 
int fun1 (int n) { int i, j, k, p, q = 0; for (i = 1; i<n; ++i) { p = 0; for (j = n; j > 1; j = j/2) ++p; for (k = 1; k < p; k = k*2) ++q; } return q; }
n^{3}  
n(log n)^{2}  
nlog n  
nlog (log n) 
for (j=n; j>1; j=j/2)  p = O(log n) ++p;
for (k=1; k
++q;
}
∴ The return value of the function fun1,
q = n log p
= n log log n
Question 32 
ABCD EFGH
 
ABCD  
HGFE DCBA
 
DCBA 
if condition fails
& returns controls
∴ DCBA will be pointed
Question 33 
51  
52  
53  
54 
f(n) = 1; if n = 1
Question 34 
2  
2  
1  
15 
When next time stkFunc (0,5) is called then, inside Switch(opcode), the control will go to Case0, where A[0]=5 and stkTop = 0+1 =1.
When next time stkFunc (0,10) is called then, inside Switch (opcode), the control will go to Case '0', where A[1]=10 and stkTop=1+1=2.
When next time stkFunc(1,0) is called from inside the printf statement, then inside Switch(opcode), the control will go to default and stkTop = 21 = 1 and value of A[1] will get returned, i.e., 10.
When next time stkFunc(1,0) is called from inside the printf statement, then inside Switch(opcode), the control will go to default and stkTop = 11 = 0 and value of A[0] will get returned, i.e., 5.
Finally the two values 10 & 5 will be added and printed.
Question 35 
# include <stdio.h> int main( ) { char s1[7] = "1234" , *p; p = s1 + 2; *p = '0' ; printf ( "%s" , s1); } 
12  
120400  
1204  
1034 
p now points to third element in s1, i.e., '3'.
*p = '0', will make value of '3' as '0' in s1. And finally s1 will become 1204.
Question 36 
void
get (
int
n)
{
if
(n < 1)
return
;
get(n1);
get(n3);
printf
(
"%d"
, n);
}
15  
25  
35  
45 
Question 37 
# include
int
main( )
{
static
int
a[] = {10, 20, 30, 40, 50};
static
int
*p[] = {a, a+3, a+4, a+1, a+2};
int
**ptr = p;
ptr++;
printf
(
"%d%d"
, ptr  p, **ptr};
}
140  
150  
160  
170 
**ptr = 40
∴ printf (“%d%d”, p + r – p, p + r) will print 140.
Question 38 
# include <stdio.h> int main( ) { int i, j, k = 0; j = 2 * 3 / 4 + 2.0 / 5 + 8 / 5; k = j; for (i = 0; i < 5; i++) { switch (i + k) { case 1: case 2: printf ( "n%d" , i + k); case 3: printf ( "n%d" , i + k); default : printf ( "n%d" , i + k); } } return 0; } 
10  
11  
12  
13 
= 6 / 4+2.0 / 5+1;
= 1 + 0.4 + 1
= 2.4
But since j is integer,
j=2
Now,
k = k  (j)
k = 0  (1) = 1
When i=0, i+k = 1,
printf executed 1 time
When i=1, i+k = 0,
printf executed 1 time
When i=2, i+k = 1,
printf executed 3 times
When i=3, i+k = 2,
printf executed 3 times
When i=4, i+k = 3,
printf executed 2 times
∴ Total no. of times printf executed is,
1 + 1 + 3 + 3 + 2 = 10
Question 39 
# include <stdio.h> int f1( void ); int f2( void ); int f3( void ); int x = 10; int main() { int x = 1; x += f1() + f2() + f3() + f2(); pirntf( "%d" , x); return 0; } int f1() { int x = 25; x++; return x; } int f2( ) { static int x = 50; x++; return x; } int f3( ) { x *= 10; return x; } 
230  
240  
250  
260 
f1( ) = 25 + 1 = 26
f2( ) = 50 + 1 = 51
f3( ) = 10 × 10 = 100
f2( ) = 51 × 1 = 52 (Since here x is static variable so old value retains)
∴ x = 1+26+51+100+52 = 230
Question 40 
#include <stdio.h> main() { int i; int *pi = &i; scanf ( "%d" , pi); printf ( "%dn" , i+5); } 
Compilation fails.  
Execution results in a runtime error.  
On execution, the value printed is 5 more than the address of variable i.  
On execution, the value printed is 5 more than the integer value entered. 
int *pi = &i; // pi is a pointer which stores the address of i.
scanf (pi); // pi = &i (we rewrite the garbage value with our values) say x = 2
printf (i+5); // i+5 = x+5 = 2+5 = 7
Hence on execution, the value printed is 5 more than the integer value entered.
Question 41 
9  
10  
11  
12 
Shift left of 1, which means the number gets doubled.
In program, shift right of 1 is given, means every time we enter the loop the number will get halved, and the value of count will get incremented by 1. And when the value of num will become zero then the while loop will get terminated. So,
num = 435/2 = 217/2 = 108/2 = 54/2 = 27/2= 13/2 = 6/2 = 3/2 = 1/2 = 0
Count = 9
So, the value count that will get returned is 9.
Question 42 
p = n * (n1) * (n2) / 6;  
p = n * (n1) / 2 * (n2) / 3;  
p = n * (n1) / 3 * (n2) / 2;  
p = n * (n1) * (n2) / 6.0; 
From the options n*(n1)*(n2) will go out of range. So eliminate A & D.
n*(n1) is always an even number. So subexpression n(n1)/2 also an even number.
n*(n1)/ 2*(n2), gives a number which is a multiple of 3. So dividing with 3 will not have any loss. Hence B is option.
Question 43 
double f( double x){ if ( abs (x*x  3) < 0.01) return x; else return f(x/2 + 1.5/x); } 
1.73  
1.74  
1.75  
1.76 
x^{2}  3 < 0.01 will become True.
So, x^{2}  3 < 0.01
x^{2}  3 < 0.01
x^{2} < 3.01
x < 1.732
Hence, x = 1.73.
Question 44 
int f( int j) { static int i = 50; int k; if (i == j) { printf (“something”); k = f(i); return 0; } else return 0; } 
The function returns 0 for all values of j.  
The function prints the string something for all values of j.  
The function returns 0 when j = 50.  
The function will exhaust the runtime stack or run into an infinite loop when j = 50. 
int f(int j)
{
static int i = 50;
int k;
if (i == j) // This will be True.
{
printf ("Something");
k = f(i); // Again called f(i) with value of i as 50. So, the function will run into infinite loop.
return 0;
}
else return 0;
}
Question 45 
for i = 1 to n do for j = 1 to n do 
output (A [i][j]); 
The matrix A itself  
Transpose of the matrix A  
Adding 100 to the upper diagonal elements and subtracting 100 from lower diagonal elements of A  
None of the above 
For first row iteration, it get swapped and becomes
For second row iteration, it comes to the original position
=A
So, it is the same matrix A.
Question 46 
7  
8  
9  
10 
P(X)=x^{5}+4x^{3}+6x+5
=x(x^{4}+4x^{2}+6)+5
=x(x(x^{3}+4x)+6)+5
=x(x(x(x^{2}+4))+6)+5
=x(x(x(x(x)+4))+6)+5
4 multiplications & 3 additions.
4 + 3 = 7
Question 47 
int unknown( int n) { int i, j, k = 0; for (i = n/2; i <= n; i++) for (j = 2; j <= n; j = j * 2) k = k + n/2; return k; } 
Θ(n^{2})  
Θ(n^{2} log n)  
Θ(n^{3})  
Θ(n^{3} logn) 
So, the total number of times loop runs is (n/2 logn).
So, the final k value will be n/2*(n/2 logn) = O(n^{2}logn)
= (n/2+1).n/2 ∙log n
= (n^{2}log n)
Question 48 
L_{2} is contextfree.  
L_{1}∩ L_{2} is contextfree.  
Complement of L_{2} is recursive.  
Complement of L_{1} is contextfree but not regular. 
Intersection of a regular language with a context free language is context free language (by closure properties of regular language) So L_{1}∩ L_{2} is contextfree.
L_{2} is CFL and CFL is not closed under complementation so we have to assume L_{2} as CSL (since every CFL is CSL) and CSL is closed under complement so Complement of L_{2} must be CSL and every CSL is recursive so Complement of L_{2} is recursive.
Since L_{1} is regular and regular language is closed under complement so complement of L_{1} must be regular and hence the statement Complement of L_{1} is contextfree but not regular is a false statement.
Question 49 
The procedure given below is required to find and replace certain characters inside an input character string supplied in array A. The characters to be replaced are supplied in array oldc, while their respective replacement characters are supplied in array newc. Array A has a fixed length of five characters, while arrays oldc and newc contain three characters each. However, the procedure is flawed
void find_and_replace(char *A, char *oldc, char *newc) { for (int i = 0; i < 5; i++) for (int j = 0; j < 3; j++) if (A[i] == oldc[j]) A[i] = newc[j]; }
The procedure is tested with the following four test cases (1) oldc = "abc", newc = "dab" (2) oldc = "cde", newc = "bcd" (3) oldc = "bca", newc = "cda" (4) oldc = "abc", newc = "bac" The tester now tests the program on all input strings of length five consisting of characters ‘a’, ‘b’, ‘c’, ‘d’ and ‘e’ with duplicates allowed. If the tester carries out this testing with the four test cases given above, how many test cases will be able to capture the flaw?
None  
2 only  
3 and 4 only  
4 only 
1, 2 works fine, 3, 4 carries flaw.
Question 50 
char inchar = 'A' ; switch (inchar) { case 'A' : printf ( "choice A n" ) ; case 'B' : printf ( "choice B " ) ; case 'C' : case 'D' : case 'E' : default : printf ( "No Choice" ) ; } 
No Choice  
Choice A  
Program gives no output as it is erroneous 
So,
→ Choice A
→ Choice B. No choice. Is the output.
Question 52 
int a, b, c = 0; void prtFun ( void ); int main () { static int a = 1; /* line 1 */ prtFun(); a += 1; prtFun(); printf ( "n %d %d " , a, b) ; } void prtFun ( void ) { static int a = 2; /* line 2 */ int b = 1; a += ++b; printf ( " n %d %d " , a, b); } What output will be generated by the given code segment?

Line 2 replaced by register int a=2;
In main there will be no change if it is static or auto because of a+=1 the auto variable a is updated to 2 from 1.
In prtfun ( ), register makes a difference.
For first print statement a is updated to 4 & prints 4, 2.
But now it is not a static variable to retain the value of a to 4. So it becomes 2, when second function call takes place & prints 4, 2 again. There is no change in b, it acts like a local variable.
Hence,
4 2
4 2
2 0.
Question 53 
char c[] = "GATE2011" ; char *p =c; printf ( "%s" , p + p[3]  p[1]) ; 
GATE2011  
E2011  
2011  
011 
p[3]  p[1] = 4, and p+4 will be pointing to the fifth position in the array 'c'. So printf starts printing from 2 and prints 2011.
Question 54 
9  
8  
5  
2 
∴ 1+0+0+0+0+0+0+0+0+1+0 = 2
Question 55 
unsigned int foo(unsigned int n, unsigned int r) { if (n > 0) return (n%r + foo (n/r, r )); else return 0; } 
345  
12  
5  
3 
∴ 5+4+3+0 = 12
Question 56 
#include void f( int *p, int *q) { p = q; *p = 2; } int i = 0, j = 1; int main() { f(&i, &j); printf ( "%d %d n" , i, j); getchar (); return 0; } 
2 2  
2 1  
0 1  
0 2 
Both pointers points to j = 1
now *p = 2
where j is updated with value 2.
printf (i, j) i = 0, j = 2
Question 57 
#include<stdio.h>
int
f(
int
*a,
int
n)
{
if
(n <= 0)
return
0;
else
if
(*a % 2 == 0)
return
*a + f(a+1, n1);
else
return
*a  f(a+1, n1);
}
int
main()
{
int
a[] = {12, 7, 13, 4, 11, 6};
printf
(
"%d"
, f(a, 6));
getchar
();
return
0;
}
9  
5  
15  
19 
if (n <= 0)
return 0;
else if (*a % 2 = = 0)
return *a + f(a+1, n1);
else
return *a – f(a+1, n1);
⇒12+(7(13(4+(11(6)))))
⇒12+(7(13(4+5)))
⇒12+7(4)
⇒12+3
⇒15
Question 58 
6  
8  
14  
15 
printf(fun(5,&x));
The code is implemented using Tail Recursion.
fun(5,15)
↓
fun(4,15)
↓
fun(3,15)
↓
fun(2,15)
↓
fun(1,15)
→ First we will trace fun(1,15) which returns 1.
→ Then trace fun(2,15) using fun(1,15) value, it returns 2.
→ Then fun(3,15), it returns 3≃(1+2)
→ Then fun(4,15), it returns 5=(2+3)
→ Then fun(5,15), it returns 8=(3+5)
If you call fun(6,15) then it will return 13=(5+8)
Here fun(n,*x)≃fun(n1,&x)+fun(n2,&x), where fun(n1,&x) is storing in variable ‘t’ & fun(n2,&x) is storing in variable x(*fp).
∴ The program is nth Fibonacci number.
Question 59 
x = 3, y = 4, z = 2  
x = 6, y = 5, z = 3
 
x = 6, y = 3, z = 5
 
x = 5, y = 4, z = 5 
→ We can directly eliminate the options B & C, because none of the variable can assign a value 4.
→ Given explanation is
a=(x>y)?((x>z)?x:z):((y>z)?y:z)
Option A:
x=3; y=4; z=2
a=(3>4)?⇒No
Then evaluate second expression⇒(4>2)?Yes
⇒a=y
a=4 (True)
Option D:
x=5; y=4; z=5
a=(5>4)⇒Yes
Then evaluate first expression ⇒ (5>5)? No
⇒ a=z ⇒ a=5 (Not true)
⇒Answer is Option A.
Question 60 
18  
19  
21  
22 
f(c,&c,&(&c))=f(4,4,4)
c is 4, b is a pointer pointing address of a, a is a pointer to pointer of c. Hence both b and c are pointing to same memory address i.e., a.
Hence whatever increment operation happens in f, it happens/ reflects on same value i.e., a.
**ppz+=1;
z=**ppz; //z=5
These steps update it to 5 and stored in z.
*py+=2; //changes c to 7, x is unchanged.
y=*py; //y=7
It updates to 7 and stored in y.
x+=3 //x is incremented by 3.
returns x+y+z=7+7+5=19
Question 61 
?1 is (getchar( ) != ’\n’) ?2 is getchar(c);  
?1 is (c = getchar( ) ) != ’\n’) ?2 is getchar(c);  
?1 is (c != ’\n’) ?2 is putchar(c);  
?1 is ((c = getchar()) != ’\n’) ?2 is putchar(c); 
{
int c;
if(?1) reverse( );
?2
}
main( )
{
printf(“Enter Text”);
printf(“\n”);
reverse( );
printf(“\n”);
}
We can simply eliminate A & B for ?2.
& Hence
?1 is ((c=getchar( )) != ‘\n’)
?2 is putchar(c);
Question 62 
1661 and 1640  
59 and 59  
1640 and 1640
 
1640 and 1661

f1(2) = 2*f1(1) + 3*f1(0) = 2
f1(3) = 2*f1(2) + 3*f1(1) = 2*2 + 3*1 = 7
f1(4) = 2*f1(3) + 3*f1(2) = 2*7 + 3*2 = 20
f1(5) = 2*f1(4) + 3*f1(3) = 2*20 + 3*7 = 40 + 21 = 61
We can skip after this as the only remaining choice is (C).
f1(6) = 2*f1(5) + 3*f1(4) = 2*61 + 3*20 = 122 + 60 = 182
f1(7) = 2*f1(6) + 3*f1(5) = 2*182 + 3*61 = 364 + 183 = 547
f1(8) = 2*f1(7) + 3*f1(6) = 2*547 + 3*182 = 1094 + 546 = 1640
Question 63 
Ic, IId, IIIb, IVa  
Ia, IId, IIIc, IVb  
Id, IIc, IIIb, IVa  
Ic, IId, IIIa, IVb

Question 64 
# include <stdio.h> int main () { char a [6] = "world" ; int i, j; for (i = 0, j = 5; i < j; a [i++] = a [j]); printf ( "%sn" , a); } /* Add code here. Remove these lines if not writing code */ 
dlrow  
Null String  
dlrld  
worow

Question 65 
# include <stdio.h> # define swapl (a, b) tmp = a; a = b; b = tmp void swap2 ( int a, int b) { int tmp; tmp = a; a = b; b = tmp; } void swap3 ( int *a, int *b) { int tmp; tmp = *a; *a = *b; *b = tmp; } int main () { int num1 = 5, num2 = 4, tmp; if (num1 < num2) {swap1 (num1, num2);} if (num1 < num2) {swap2 (num1 + 1, num2);} if (num1 >= num2) {swap3 (&num1, &num2);} printf ( "%d, %d" , num1, num2); } /* Add code here. Remove these lines if not writing code */ 
5, 5  
5, 4  
4, 5  
4, 4 
{Swap3 (&num1, &num2) ; }"
Statement is true, so call by reference will be performed and the value of num1 and num2 will get exchanged.
Question 66 
#include <stdio.h> int main () { int i, j; int a [8] = {1, 2, 3, 4, 5, 6, 7, 8}; for (i = 0; i < 3; i++) { a[i] = a[i] + 1; i++; } i; for (j = 7; j > 4; j) { int i = j/2; a[i] = a[i]  1; } printf ( "%d, %d" , i, a[i]); } /* Add code here. Remove these lines if not writing code */ 
2, 3  
2, 4  
3, 2  
3, 3 
First for loop will run for i = 0, 2 and 4 as i is incremented twice and resultant array will be 'a' = 2, 2, 4, 4, 5, 6, 7, 8. Loop will terminate at i = 4.
After that value of 'i' will become '3' as there is a decrement operation after for loop.
Next for loop is running for j = 7, 6, 5 and corresponding 'i' values which is a local variable inside for loop will be
3(7/2), 3(6/2), 2(5/2)
Array after this for loop will be
a = 2, 2, 3, 2, 5, 6, 2, 8.
After the loop, current 'i' value is 3 and elements at a[3] = 2.
Question 67 
# include <stdio.h> int main () { int i, j; char a [2] [3] = {{ 'a' , 'b' , 'c' }, { 'd' , 'e' , 'f' }}; char b [3] [2]; char *p = *b; for (i = 0; i < 2; i++) { for (j = 0; j < 3; j++) { *(p + 2*j + i) = a [i] [j]; } } } /* Add code here. Remove these lines if not writing code */ 
*(p+2) = a[0][1] = b
*(p+4) = a[0][2] = c
*(p+1) = a[1][0] = d
*(p+3) = a[1][1] = e
*(p+5) = a[1][2] = f
Question 68 
void f ( int n) { if (n <=1) { printf ( "%d" , n); } else { f (n/2); printf ( "%d" , n%2); } } 
010110101  
010101101  
10110101  
10101101 
So, after traversing the tree we get:
10101101
Question 69 
void f ( int n) { if (n <= 1) { printf ( "%d" , n); } else { f (n/2); printf ( "%d" , n%2); } } 
void f ( int n) { if (n/2) { f(n/2); } printf ( "%d" , n%2); } 
void f ( int n) { if (n <=1) { printf ( "%d" , n); } else { printf ( "%d" , n%2); f (n/2); } } 
Both P1 and P2  
P2 only  
P1 only  
Neither P1 nor P2 
But P2 prints the reverse of original sequence printed by original program.
Question 70 
#include <stdio.h> int f( int n) { static int r = 0; if (n <= 0) return 1; if (n > 3) { r = n; return f(n2)+2; } return f(n1)+r; } int main() { printf ( "%d" , f(5)); } 
5  
7  
9  
18 
f(5) = f(3) + 2
f(3) = f(2) + 5 (where r is static and value of r=5)
f(2) = f(1) + 5
f(1) = f(0) + 5
f(0) = 1
⟹ f(5) = 1+5+5+5+2 = 18
Question 71 
int f (int n) { if (n <= 1) return 1; else if (n % 2 == 0) return f(n/2); else return f(3n  1); }Assuming that arbitrarily large integers can be passed as a parameter to the function, consider the following statements. 1. The function f terminates for finitely many different values of n ≥ 1. ii. The function f terminates for infinitely many different values of n ≥ 1. iii. The function f does not terminate for finitely many different values of n ≥ 1. iv. The function f does not terminate for infinitely many different values of n ≥ 1. Which one of the following options is true of the above?
(i) and (iii)  
(i) and (iv)  
(ii) and (iii)  
(ii) and (iv) 
→ Let n=3, then it is terminated in 2^{nd} iteration.
→ Let n=5, then sequence is 5→14→7→20→10 and it will repeat.
→ Any number with factor 5 and 2 leads to infinite recursion.
So, (iv) is True and (iii) is False.
Question 72 
#include <stdio.h> int main () { int sum = 0, maxsum = 0, i, n = 6; int a [] = {2, 2, 1, 3, 4, 2}; for (i = 0; i < n; i++) { if (i == 0  a [i] < 0  a [i] < a [i  1]) { if (sum > maxsum) maxsum = sum; sum = (a [i] > 0) ? a [i] : 0; } else sum += a [i]; } if (sum > maxsum) maxsum = sum ; printf ("%dn", maxsum); }
9  
8  
7  
6 
Question 73 
int i ; program main () { int j = 60; i = 50; call f (i, j); print i, j; } procedure f (x, y) { i = 100; x = 10; y = y + i ; }Which one of the following options represents the correct output of the program for the two parameter passing mechanisms?
Call by value : i = 70, j = 10; Call by reference : i = 60, j = 70  
Call by value : i = 50, j = 60; Call by reference : i = 50, j = 70  
Call by value : i = 10, j = 70; Call by reference : i = 100, j = 60  
Call by value : i = 100, j = 60; Call by reference : i = 10, j = 70 
'i' is a global variable. Then in main( ) a local variable 'j' as integer declared, i.e., j=60 and global varible 'i' is initialized to 50 by i=50.
Now procedure f called and values of 'i' and 'j' are passed to it, i.e., in f(i, j)→f(x, y).
Content of memory location of 'i' (here 50) is copied to memory location of x (which is different from i) and content of memory location of 'j' (here 60) is copied to memory location of y. Then in f(x, y) i=100 changes the global i to 100, x=10 changes the local x from 50 to 10 and y=y+i means y=60+100=160. Now, when return back to main, i & j will be 100 & 60 respectively.
Call by reference:
Now procedure f called and passed reference of i and j to it, i.e., in f(i,j)→f(x,y). x and y are pointing to same memory location of i and j respectively. So i=100 changes the global i to 100 and x=10 means as well as global i=10 (as the i being passed is the global variable and x and i share the same address).
y=y+i means y=60+10=70 and this changes the value of j also to as j and y have the same addresses. Now when return to main, i and j will be 10 and 70.
Question 74 
int i ; program main () { i = 10; call f(); } procedure f() { int i = 20; call g (); } procedure g () { print i; }Let x be the value printed under static scoping and y be the value printed under dynamic scoping. Then, x and y are
x = 10, y = 10  
x = 20, y = 10  
x = 10, y = 20  
x = 20, y = 20

Question 75 
Early, late, decrease, increase  
Late, early, increase, decrease  
Late, early, decrease, increase  
Early, late, increase, decrease 
Dynamic scoping requires late binding (during execution time).
Late binding decreases efficiency as this binding needs to be done at run time.
Question 76 
void swap( int *px, int *py) { *px = *px  *py; *py = *px + *py; *px = *py  *px; } 
S1  
S2 and S3  
S2 and S4
 
S2 and S5

We may get the segmentation fault if the pointer values are constant (i.e., px or py) (or) (px or py) are points to a memory location is invalid.
S4:
Swap procedure can be implemented correctly but not for all input pointers because arithmetic overflow may occur based on input values.
Question 77 
subroutine swap(ix,iy) it = ix L1 : ix = iy L2 : iy = it end ia = 3 ib = 8 call swap (ia, 1b+5) print *, ia, ib endS1: The compiler will generate code to allocate a temporary nameless cell, initialize it to 13, and pass the address of the cell swap S2: On execution the code will generate a runtime error on line L1 S3: On execution the code will generate a runtime error on line L2 S4: The program will print 13 and 8 S5: The program will print 13 and 2 Exactly the following set of statement(s) is correct:
S1 and S2
 
S1 and S4
 
S3  
S1 and S5

Swap (8, 13)
⇒ ia will returns value with 13.
⇒ ib is unchanged, because here we using pass by reference value.
➝ Temporary nameless is initialized to 13.
➝ There is No runtime error.
Question 78 
#include <stdio.h> struct test { int i; char *c; }st[] = {5, "become", 4, "better", 6, "jungle", 8, "ancestor", 7, "brother"}; main () { struct test *p = st; p += 1; ++p > c; printf("%s,", p++ > c); printf("%c,", *++p > c); printf("%d,", p[0].i); printf("%s n", p > c); }
jungle, n, 8, ncestor  
etter, u, 6, ungle  
cetter, k, 6, jungle  
etter, u, 8, ncestor 
Line 1  main ( )
Line 2  {
Line 3  struct test *p = st;
Line 4  p += 1;
Line 5  ++p → c;
Line 6  printf("%s", p++→ c);
Line 7  printf("%c", +++p → c);
Line 8  printf("%d", p[0].i);
Line 9  printf("%s\n", p → c);
Line 10  }
Now,
Line 3: Initially p is pointing to st, i.e., first element of st which is {5, "become"}
Line 4: Now p is pointing to {4, "better"}
Line 5: ++(p → c), since → has higher precedence, so p → c points to 'e' of "better".
Line 6: prints 'enter' and p now points to {6, "jungle"}
Line 7: ***(p → c), since → has higher precedence. So, prints 'u'.
Line 8: p → i, which is 6 so prints '6'.
Line 9: prints 'ungle' since p is pointing to 'u'.
So, output is "enter, u, 6, ungle".
Question 79 
#include void swap (int *x, int *y) { static int *temp; temp = x; x = y; y = temp; } void printab () { static int i, a = 3, b = 6; i = 0; while (i <= 4) { if ((i++)%2 == 1) continue; a = a + i; b = b + i; } swap (&a, &b); printf("a = %d, b = %dn", a, b); } main() { printab(); printab(); }
a = 0, b = 3 a = 0, b = 3  
a = 3, b = 0 a = 12, b = 9  
a = 3, b = 6 a = 3, b = 6  
a = 6, b = 3 a = 15, b = 12 
Inside print 'a' and 'b' are added to odd integers from 1 to 5, i.e., 1+3+5=9. So, in first call to print ab,
a = 3+9 = 6
b = 6+9 = 3
Static variable have one memory throughout the program run (initialized during program start) and they keep their values across function calls. So during second call to print ab,
a = 6+9 = 15
b = 3+9 = 12
Question 80 
#include int a1[] = {6, 7, 8, 18, 34, 67}; int a2[] = {23, 56, 28, 29}; int a3[] = {12, 27, 31}; int *x[] = {a1, a2, a3}; void print(int *a[]) { printf("%d,", a[0][2]); printf("%d,", *a[2]); printf("%d,", *++a[0]); printf("%d,", *(++a)[0]); printf("%dn", a[1][+1]); } main() { print(x); }
8, 12, 7, 23, 8  
8, 8, 7, 23, 7  
12, 12, 27, 31, 23  
12, 12, 27, 31, 56 
It returns the value of 3^{rd} element in a1.
First printf print 8.
2) *a[2] = *(*(a+2))
It returns the value of 1^{st} element in a3.
Second printf print 12.
3) *++a[0] = *(++(*(a+0)))
a[0] is pointing to 1^{st} element in a1.
++a[0]  after preincrement performed, now a[0] is pointing to 2^{nd} element in a1.
*++a[0] return the value of 2^{nd} element in a1.
Third printf print 7.
4) *(++a)[0]
++a  after preincrement is performed 'a' is pointing to a2.
(++a)[0] is pointing to 1^{st} element in a2.
*(++a)[0] returns the value of 1^{st} element in a2.
Fourth printf print 23.
5) a[1][+1] = *(*(a1)+1)
(a1) is pointing to a1.
*(a1) is pointing to the 2^{nd} element in a1, because in 3^{rd} printf already a1 was incremented by 1.
*(a1)+1 is pointing 3^{rd} element in a1.
*(*(a1)+1) returns the value of 3^{rd} element in a1, i.e., 8.
Question 81 
int func(int m, int n) { if (E) return 1; else return(func(m 1, n) + func(m  1, n  1)); }In the above function, which of the following is the correct expression for E?
(n == 0)  (m == 1)  
(n == 0) && (m == 1)  
(n == 0)  (m == n)  
(n == 0) && (m == n) 
^{m}C_{0} = 1
^{n}C_{n} = 1
Question 82 
(i)  B, (ii)  D, (iii)  E, (iv)  F, (v)  G, (vi)  A  
(i)  C, (ii)  A, (iii)  E, (iv)  D, (v)  H, (vi)  F  
(i)  C, (ii)  F, (iii)  H, (iv)  A, (v)  G, (vi)  D  
(i)  B, (ii)  E, (iii)  C, (iv)  F, (v)  G, (vi)  H 
Question 83 
int
( * f) (
int
* ) ;
A function that takes an integer pointer as argument and returns an integer.  
A function that takes an integer as argument and returns an integer pointer.  
A pointer to a function that takes an integer pointer as argument and returns an integer.  
A function that takes an integer pointer as argument and returns a function pointer. 
→ A pointer to a function which takes integer as a pointer and return an integer value.
Question 84 
both are procedural languages
 
both are based on λcalculus
 
both are declarative
 
both use Hornclauses

Question 85 
(i) and (ii) only
 
(i) and (iv) only
 
(i), (ii) and (iv) only
 
(i), (iii) and (iv) only

Question 86 
double foo ( double ); /* Line 1 */ int main() { double da, db; // input da db = foo(da); } double foo( double a) { return a; } 
no compile warning or error  
some compilerwarnings not leading to unintended results  
some compilerwarnings due to typemismatch eventually leading to unintended results
 
compiler errors 
Question 87 
double foo ( double ); /* Line 1 */ int main() { double da, db; // input da db = foo(da); } double foo( double a) { return a; } 
8, 4, 0, 2, 14
 
8, 4, 0, 2, 0
 
2, 0, 4, 8, 14
 
2, 0, 4, 8, 0

⇒ foo (a, sum) = foo (2048,0)
⇒ n == 2048
⇒ k = n%10 = 2048%10 = 8
⇒ j = n/10 = 2048/10 = 204
Sum = Sum+k = 0+8 = 8
foo(j, sum) = foo(204, 8)
⇒ n=204
k = n%10 = 204%10 = 4
j = n/10 = 204/10 = 20
sum = sum+k = 12+0 = 12
foo(j, sum) =foo(2,12)
⇒ n=2
k = n%10 = 2%10 = 2
j = n/10 = 2/10 = 0
sum = 14
foo(0,14) ⇒ n==0
printf("%d", k) ⇒ k = 2, 0, 4, 8
After foo( ) statement one more printf statement is there then if print 0 after all digits of n.
2, 0, 4, 8, 0.
Question 88 
int anagram (char *a, char *b) { int count [128], j; for (j = 0; j < 128; j++) count[j] = 0; j = 0; while (a[j] && b[j]) { A; B; } for (j = 0; j < 128; j++) if (count [j]) return 0; return 1; }Choose the correct alternative for statements A and B.
A : count [a[j]]++ and B : count[b[j]]–  
A : count [a[j]]++ and B : count[b[j]]++  
A : count [a[j++]]++ and B : count[b[j]]–  
A : count [a[j]]++and B : count[b[j++]]– 
B: Decrements the count by 1 at each index that is equal to the ASCII value of the alphabet it is pointing at. Also it increments the loop counter for next iteration.
If one string is permutation of other, there would have been equal increments and decrements at each index of array, and so count should contain zero at each index, that is what the loop checks at last and if any nonzero elements is found, it returns 0 indicating that strings are not anagram to each other.
Question 89 
struct node { int value; struct node *next; ); void rearrange ( struct node *list) { struct node *p, *q; int temp; if (!list  !list > next) return ; p = list; q = list > next; while (q) { temp = p > value; p > value = q > value; q > value = temp; p = q > next; q = p ? p > next : 0; } } 
1, 2, 3, 4, 5, 6, 7  
2, 1, 4, 3, 6, 5, 7  
1, 3, 2, 5, 4, 7, 6  
2, 3, 4, 5, 6, 7, 1 
Question 90 
i = 0; j = 1; while (j < n ) { if (E) j++; else if (a[j]  a[i] == S) break ; else i++; } if (j < n) printf ( "yes" ) else printf ( "no" ); 
Choose the correct expression for E.
a[j] – a[i] > S  
a[j] – a[i] < S  
a[i] – a[j] < S  
a[i] – a[j] > S

If at times difference becomes greater than S we know that it won't reduce further for same 'i' and so we increment the 'i'.
Question 91 
 a[i] ≥ b [i] => c[2i] ≥ a [i]
 a[i] ≥ b [i] => c[2i] ≥ b [i]
 a[i] ≥ b [i] => c[2i] ≤ a [i]
 a[i] ≥ b [i] => c[2i] ≤ b [i]
only I and II  
only I and IV  
only II and III  
only III and IV 
Since both 'a' and 'b' are sorted in the beginning, there are 'i' elements than or equal to a[i] and similarly 'i' elements smaller than or equal to b[i]. So, a[i] ≥ b[i] means there are 2i elements smaller than or equal to a[i] and hence in the merged array, a[i] will come after these 2i elements. So, c[2i] ≤ a[i].
Similarly, a[i] ≥ b[i] says for b that, there are not more than 2i elements smaller than b[i] in the sorted array. So, b[i] ≤ c[2i].
So, option (C) is correct.
Question 92 
#include<stdio.h> int f( int n, int k) { if (n == 0) return 0; else if (n % 2) return f(n/2, 2*k) + k; else return f(n/2, 2*k)  k; } int main () { printf ( "%d" , f(20, 1)); return 0; } 
5  
8  
9  
20 
Hence, 9 is the answer.
Question 93 
call swap (x, y)
 
call swap (&x, &y)
 
swap (x,y) cannot be used as it does not return any value
 
swap (x,y) cannot be used as the parameters are passed by value

Here parameters passed by value in C then there is no change in the values.
Option B:
Here values are not swap.
Here parameters are passed by address in C.
Option C:
It is false. Return value is not valid for exchanging the variables.
Option D:
It is correct.
We cannot use swap(x,y) because parameters are passed by value.
Only exchanging the values (or) variables are passing their address and then modify the content with the help of dereferencing operator(*).
Question 94 
have well indented programs
 
be able to infer the flow of control from the compiled code
 
be able to infer the flow of control from the program text
 
avoid the use of GOTO statements

Question 95 
5  
6  
7  
8 
The value return by f(1) = 7
Question 96 
n = d_{1}d_{2}…d_{mi} and rev = d_{m}d_{m1}…d_{mi+1}
 
n = d_{mi+1}…d_{m1}d_{m} or rev = d_{mi}…d_{2}d_{1}
 
n ≠ rev  
n = d_{1}d_{2}…d_{m} and rev = d_{m}…d_{2}d_{1}

Question 97 
gnirts  
string
 
gnirt  
no output is printed

Question 98 
(i), (iv), (vi), (viii)
 
(i), (iv), (vii)  
(i), (iii), (v), (vi), (viii)
 
(ii), (v), (viii)

Question 99 
x + y using repeated subtraction
 
x mod y using repeated subtraction
 
the greatest common divisor of x and y
 
the least common multiple of x and y

Question 100 
P  2, Q  3, R  4, S  1
 
P  4, Q  3, R  2, S  1
 
P  3, Q  4, R  1, S  2
 
P  3, Q  4, R  2, S  1

Q) Logic is also declarative but involves theorem proving.
R) Object oriented is imperative statement based and have abstract data types.
S) Imperative programs are made giving commands and follow definite procedure.
Question 101 
x = 1 + x;  
x = 1  x;  
x = x  1;  
x = 1 % x; 
For x = 0, it gives 1.
For x = 1, it gives 0.
Question 102 
A program attempts to generate as many permutations as possible of the string, 'abcd' by pushing the characters a, b, c, d in the same order onto a stack, but it may pop off the top character at any time. Which one of the following strings CANNOT be generated using this program?
abcd  
dcba  
abad  
cabd 
B) First push abcd, and after that pop one by one. Sequence of popped elements will come to dcba.
C) push abc, and after that pop one by one. Sequence of popped elements will come to cba. Now push 'd' and pop 'd', final sequence comes to cbad.
D) This sequence is not possible because 'a' cannot be popped before 'b' anyhow.
Question 103 
#include<stdio.h> #define ROW 4 #define COL 4 int M[ROW][COL] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}; main() { int i, j, t; for (i = 0; i < 4; ++i) { X } for (1 = 0; i < 4; ++i) for (j = 0; j < 4; ++j) printf ("%d", M[i][j]); }
In (D) , given statements is wrong as temporary variable needs to be assigned some value and not viceversa.
Question 104 
#include <stdio.h> int funcf (int x); int funcg (int y); main() { int x = 5, y = 10, count; for (count = 1; count <= 2; ++count) { y += funcf(x) + funcg(x); printf ("%d ", y); } } funcf(int x) { int y; y = funcg(x); return (y); } funcg(int x) { static int y = 10; y += 1; return (y+x); }
43 80  
42 74  
33 37  
32 32 
In first case of funcf, which in turn calls funcg, y becomes 11 and it returns 5+11=16.
In second call of funcg, y becomes 12 and it returns 5+12=17.
So, in main y is incremented by 16+17=33 to become 10+33 =43.
In second iteration:
y will be incremented by 18+19=37 to give 43+37=80.
Question 105 
#include <stdio.h> void wrt_it (void); int main (void) { printf("Enter Text"); printf ("n"); wrt_ it(); printf ("n"); return 0; } void wrt_it (void) { int c; if (?1) wrt_it(); ?2 }
?1 is getchar() ! = ‘\n’ ?2 is getchar(c);  
?1 is (c = getchar()); ! = ‘\n’ ?2 is getchar(c);  
?1 is c! = ‘\n’ ?2 is putchar(c);  
?1 is (c = getchar()) ! = ‘\n’ ?2 is putchar(c); 
putchar( ) = writes a character specified by the argument to stdout.
As getchar( ) and putchar( ), both are needed to read the string and prints its reverse and only option (D) contains both the function. (D) is the answer.
Now coming to the code, wrt_id(void) is calling itself recursively. When \n is encountered, putchat( ) gets executed and prints the last character and then the function returns to its previous call and prints last 2^{nd} character and so on.
Question 106 
#include <stdio.h> typedef struct { char *a; char *b; } t; void f1(t s); void f2(t *p); main() { static t s = {"A", "B"}; printf ("%s %sn", s.a, s.b); f1(s); printf ("%s %sn", s.a, s.b); f2(&s); } void f1(t s) { s.a = "U"; s.b = "V"; printf ("%s %sn", s.a, s.b); return; } void f2(t *p) { p > a = "V"; p > b = "W"; printf("%s %sn", p > a, p > b); return; }What is the output generated by the program ?
AB UV VW VW  
AB UV AB VW  
AB UV UV VW  
AB UV VW UV 
→ f1 is call by value. The changes applicable only for local from f1. UV is printed.
→ Back in main( ), AB is printed.
→ Then in f2, VW is printed.
Hence, answer is (B).
Question 107 
(∀x) [(∃y) [B(x,y) ∧ C(y)] ⇒ A(x)] ∧ ¬(∃x)[B(x,x)]  
(∀x) [(∀y) [B(x,y) ∧ C(y)] ⇒ A(x)] ∧ ¬(∃x)[B(x,x)]  
(∀x) [(∃y) [B(x,y) ∧ C(y)] ⇒ A(x)] ∨ ¬(∃x)[B(x,x)]  
(∀x) [(∀y) [B(x,y) ∧ C(y)] ⇒ A(x)] ∧ (∃x)[B(x,x)] 
Question 108 
global int i = 100, j = 5; void P(x) { int i = 10; print(x + 10); i = 200; j = 20; print(x); } main() { P(i + j); } 
115, 220  
25, 220
 
25, 15  
115, 105

P(100+5) = P(105)
→void P(105)
{
int i=10;
print (x+10); ⇒ 105+10=115 prints
i=200;
j = 20;
print (x); ⇒ x=105 prints
}
115, 105 prints
Question 109 
global int i = 100, j = 5; void P(x) { int i = 10; print(x + 10); i = 200; j = 20; print(x); } main() { P(i + j); } 
115, 220  
25, 220
 
25, 15
 
115, 105

In void P(x)
{ int i = 10;
print(x + 10); ⇒ 105+10=115 prints
print (x); ⇒ print x=220;
Question 110 
Class P { void f(int i) { print(i); } } Class Q subclass of P { void f(int i) { print(2*i); } }Now consider the following program fragment:
P x = new Q(); Q y = new Q(); P z = new Q(); x.f(1); ((P)y).f(1); z.f(1);Here ((P)y) denotes a typecast of y to P. The output produced by executing the above program fragment will be
1 2 1  
2 1 1  
2 1 2
 
2 2 2 
Note: The given question is not in the present syllabus
Question 111 
for (k = 3; k < = n; k++) A[k] = 0; for (k = 2; k < = TwoLog_n; k++) for (j = k + 1; j < = n; j++) A[j] = A[j]  (j % k); for (j = 3; j < = n; j++) if (!A[j]) printf ( "%d" , j); 
{mm ≤ n, (∃i)[m=i!]}
 
{mm ≤ n, (∃i)[m=i^{2}]}
 
{mm ≤ n, m is prime}
 
{ } 
Now Trace the code,
for (k=3; k<=n; k++)
A[k]=0; // A[3]=0
A[4]=0
for (k=2; k<=Two log_n; k++)
for(j=k+1; j<=n; j++)
A[j] = A[j] // (j%k); // A[3] = 0 // I=1
A[4] = 0 // I=1
for (j=3; j<=n; j++)
if (!A[j]) printf("%d", j);
// if (!1) means if (0), so printf will never execute
Hence, Option (D) is the answer.
Question 112 
Consider the C program shown below.
#include <stdio.h> #define print(x) printf("%d ", x) int x; void Q( int z) { z += x; print(z); } void P( int *y) { int x = *y + 2; Q(x); *y = x  1; print(x); } main( void ) { x = 5; P(&x); print(x); } 
12 7 6
 
22 12 11
 
14 6 6
 
7 6 6 
p(&x) it goes to P( ) function
y=5
x=5+2=7;
Q(x)
z=7
z=7+5=12(Print+z→I)
comes to P( )
*y=71=6
x=7(Print x→II)
comes to main ( ),
print x=*y=6 (print x→III)
Output: 12 7 6
Question 113 
At most one activation record exists between the current activation record and the activation record for the main  
The number of activation records between the current activation record and the activation record for the main depends on the actual function calling sequence.  
The visibility of global variables depends on the actual function calling sequence.  
Recursion requires the activation record for the recursive function to be saved on a different stack before the recursive fraction can be called. 
Question 114 
An array, each element of which is a pointer to a structure of type node  
A structure of 2 fields, each field being a pointer to an array of 10 elements  
A structure of 3 fields: an integer, a float, and an array of 10 elements  
An array, each element of which is a structure of type node 
Question 115 
X – 1 Y – 3 Z – 2  
X – 2 Y – 1 Z – 3  
X – 3 Y – 2 Z – 1  
X – 3 Y – 1 Z – 2 
Y → n is pointer to invalid memory, a making it as a dangling pointer.
Z → p is not initialized.
p = malloc (size of(char))p = malloc (size of(char)); should have been used before assigning 'aa' to ∗p.
Question 116 
multiple variables having the same memory location  
multiple variables having the same value  
multiple variables having the same identifier  
multiple uses of the same variable 
Question 117 
22 bytes  
14 bytes  
18 bytes  
10 bytes 
max[float, long] = max [4, 8] = 8
Total = short[5] + max[float,long] = 10 + 8 = 18
Question 118 
10  
4  
6  
7 
i=1; count=1
i=2; count=3
i=3; count=6
i=4; count=10
It return count value is 10.
Question 119 
Theory Explanation is given below. 
Question 120 
Given the programming constructs (i) assignment (ii) for loops where the loop parameter cannot be changed within the loop (iii) ifthenelse (iv) forward go to (v) arbitrary go to (vi) nonrecursive procedure call (vii) recursive procedure/function call (viii) repeat loop, which constructs will you not include in a programming language such that it should be possible to program the terminates (i.e., halting) function in the same programming language.
(ii), (iii), (iv)  
(v), (vii), (viii)  
(vi), (vii), (viii)  
(iii), (vii), (viii) 
Question 121 
Finds the maximum of a, b, and c  
Finds the minimum of a, b and c  
Finds the middle number of a, b, c  
None of the above

Question 122 
89  
90  
91  
92 
fun(95) = fun(fun(106))
= fun(96)
= fun(fun(107))
= fun(97)
= fun(fun(108))
= fun(98)
= fun(fun(109))
= fun(99)
= fun(110)
= fun(100)
= fun(fun(111))
= fun(101)
= 91
Question 123 
5  
25  
36  
42 
If it is call by value then answer is 36.
Question 124 
x or A, y, x of B and z in S1 and x of B, y and i in S2
 
x or B, y and z in S1 and x of B, i and z in S2  
x or B, z and y in S1 and x of A, i and y in S2
 
None of the above 
Question 125 
10  
20  
10  
None 
X = 10
Question 126 
I  
II  
III  
All of the above 
Question 127 
Computes the LCM of two numbers  
Divides the larger number by the smaller number  
Computes the GCD of two numbers  
None of the above 
1^{st} pass : X=3 and Y=2
2^{nd} pass : X=1 and Y=2
3^{rd} pass : X=1 and Y=1
Write(X), which writes 1. Which is nothing but GCD of 3 & 5.
Question 128 
2  
Run time error  
None of the above 
X in the procedure FIND is a local variable. No change will be reflected in global variable X.
Question 129 
2800  
2400  
2000  
1200 
Question 130 
1, because m is a local variable in P  
0, because m is the actual parameter that corresponds to the formal parameter in p
 
0, because both x and y are just reference to m, and y has the value 0  
1, because both x and y are just references to m which gets modified in procedure P  
none of the above 
Question 131 
0, because n is the actual parameter corresponding to x in procedure Q.  
0, because n is the actual parameter to y in procedure Q.  
1, because n is the actual parameter corresponding to x in procedure Q.  
1, because n is the actual parameter corresponding to y in procedure Q.  
none of the above 
Question 132 
PARAM, P, Q  
PARAM, P  
PARAM, Q  
P, Q  
none of the above 
Question 133 
exchanges a and b  
doubles a and stores in b  
doubles b and stores in a  
leaves a and b unchanged  
none of the above 
Let us consider a=5; b=2
a := a+b = 5+2 = 7
b := ab = 72 = 5
a := ab = 75 = 2
O/P: a=2; b=5
Question 134 
10  
11  
3  
None of the above 
W(n)=W(3)
Procedure W(var x; int)
begin
x = x+1 = 3+1 = 4
Print x → Print x=4
end
Question 135 
10, 3  
31, 3  
27, 7  
None of the above 
And variable y and z of func1 points to address of variable x.
Therefore, y = y+4 ⇒ y = 10+4 = 14
and z = x+y+z ⇒ z = 14+14+3 = 31
z will be stored back in k.
Hence, x=31 and y will remain as it is (y=3).
Hence, answer is (B).
Question 138 
41 
T(n) = T(n1) + T(n2) + 2
T(0) = T(1) = 0 (for fib(0) and fib(1), there are no extra recursive calls)
T(2) = 2
T(3) = 4
T(4) = 8
T(5) = 14
T(6) = 24
T(7) = 40
Counting the initial call, we get
40+1 = 41
Question 139 
None of the above 
Question 140 
(a)  (s), (b)  (r), (c)  (p), (d)  (q) 
Question 141 
Passing an expression as a parameter  
Passing an array as a parameter  
Passing a pointer as a parameter  
Passing as array element as a parameter  
Both A and D 
Passing array element as a parameter.
Consider an example:
{
.......
a[ ] = {1, 2, 3, 4, 5}
fun (a[i]);
print a[0];
}
fun (int x)
{
int i=1;
}
O/P:
Callbyreference: 6
Callbyname: 1
Result is different.
Option A:
While we passing an expression as a parameter due to precedence (higher (or) lower), the output may changes.
If we pass 1+2 to the below function
int foo (int x)
{
return x*x;
}
O/P:
Call by reference = 3*3 = 9
Call by name = 1+2*1+2 (* has higher precedenc e)
= 1+2+2
= 5
Output differs.
Answer: A, D
Question 142 
It makes it more difficult to verify programs.  
It makes programs more inefficient.  
It makes it more difficult to modify existing programs.  
It results in the compiler generating longer machine code. 
Question 143 
3, 6  
6, 7  
3, 7  
None of the above. 
Question 144 
Var x,y :integer; begin x:=(n+2)/(n3) end; procedure Q Var x,y :integer; begin x:=3; y:=4; P(y); write (x)(1) end; begin x:=7; y:=8; Q: Write (x);(2) end.
3, 6  
6, 7  
3, 7  
None of the above 