PRogramming
Question 2 
0, c  
0, a+2  
‘0’, ‘a+2’  
‘0’, ‘c’ 
Question 2 Explanation:
char x = ‘a’+2;
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.
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 3 
Hi Bye Bye Hi  
Hi Bye Hi Bye  
Bye Hi Hi Bye  
Bye Hi Bye Hi 
Question 3 Explanation:
The first call to the function ‘func1(str1, str2);’ is call by value. 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”.
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 4 
4  
5  
6  
40 
Question 4 Explanation:
Since 240 is the input, so first loop will make j=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.
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 5 
10230  
10231  
10232  
10233 
Question 5 Explanation:
#include
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
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 6 
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 
Question 6 Explanation:
Option A: 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.
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 7 
53423122233445  
53423120112233  
53423122132435  
53423120213243 
Question 7 Explanation:
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 8 
Return of 6 and 6 respectively.  
Infinite loop and abnormal termination respectively.  
Abnormal termination and infinite loop respectively.  
Both terminating abnormally. 
Question 8 Explanation:
while(val>0)
{ 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.
{ 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 9 
3  
4  
5  
6 
Question 9 Explanation:
main( )
{
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)=35=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.
{
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)=35=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 10 
23  
24  
25  
26 
Question 10 Explanation:
Arithmetic operators have least priority in this case, as count+=v & 1, we first compute v& 1 and then adds to count variable.
Question 11 
0, 0  
0, 1  
1, 0  
1, 1 
Question 11 Explanation:
printxy (int x, int y)
{
int *ptr;
x = 0;
ptr = &x;
y = *ptr;
*ptr = 1;
}
printxy (1, 1)
{
int *ptr;
x = 0;
ptr = &x;
y = *ptr;
*ptr = 1;
}
printxy (1, 1)
Question 12 
(q==r) && (r==0)  
(x>0) && (r== x) && (y>0)  
(q==0) && (r==x) && (y>0)  
(q==0) && (y>0) 
Question 12 Explanation:
Divide x by y.
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).
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 13 
3  
4  
5  
6 
Question 13 Explanation:
Swap (&x, &y) exchanges the contents of x & y.
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
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 14 
0  
1  
2  
3 
Question 14 Explanation:
n=++m; (pre)
n=11
n1=m++ (post)
Here, n1=11, but m will be updated to 12.
n & n1 decrements 11 to 10 & subtracting gives us zero.
Question 15 
1  
2  
4  
6 
Question 15 Explanation:
char * C = “GATECSIT2017”;
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
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 16 
f(s,*s)  
i= f(i,s)  
f(i,*s)  
f(i,*p) 
Question 16 Explanation:
int i = 100;
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
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 17 
2016  
2017  
2018  
2019 
Question 17 Explanation:
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 18 
a != n  
b != 0  
b > (a + 1)  
b != a 
Question 18 Explanation:
main ( )
{
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.
{
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 19 
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 
Question 19 Explanation:
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 20 
6, 2  
6, 6  
4, 2  
4, 4 
Question 20 Explanation:
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 21 
30  
31  
32  
33 
Question 21 Explanation:
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 22 
X^{Y} =a^{b}  
(res*a)^{Y} = (res*X)^{b}  
X^{Y} =res*a^{b}  
X^{Y} = (res*a)^{b} 
Question 22 Explanation:
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 = 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})
{
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 23 
3  
4  
5  
6 
Question 23 Explanation:
Given
f(a, 5) ⇒ f(100, 5)
f(a, 5) ⇒ f(100, 5)
Question 24 
5  
6  
7  
8 
Question 24 Explanation:
Function f_{1} will not swap the value of 'a' and 'b' because f_{1} is call by value.
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
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 25 
2036, 2036, 2036  
2012, 4, 2204  
2036, 10, 10  
2012, 4, 6 
Question 25 Explanation:
⇒ Address of x = 2000
⇒ 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
⇒ 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 26 
{r = qx+y ∧ r  
{x = qy+r ∧ r  
{y = qx+r ∧ 0  
{q+1 
Question 26 Explanation:
The loop terminates when r
In each iteration q is incremented by 1 and y is subtracted from 'r'. Initial value of 'r' is 'x'. So, loop iterates x/y times and q will be equal to x/y and r = x%y.
⇒ x = qy + r
⇒ x = qy + r
Question 27 
5  
6  
7  
8 
Question 27 Explanation:
Note: Out of syllabus.
Question 28 
n^{3}  
n(log n) ^{2}  
nlog n  
nlog (log n) 
Question 28 Explanation:
int fun1 (int n) {
int i, j, k, p, q = 0;
for (i=1; i
return q;
}
∴ nlog(logn)
int i, j, k, p, q = 0;
for (i=1; i
return q;
}
∴ nlog(logn)
Question 29 
ABCD EFGH
 
ABCD  
HGFE DCBA
 
DCBA 
Question 29 Explanation:
if condition fails
& returns controls
∴ DCBA will be pointed
Question 30 
51  
52  
53  
54 
Question 30 Explanation:
Recurrence Relation is
f(n) = 1; if n = 1
f(n) = 1; if n = 1
Question 31 
2  
2  
1  
15 
Question 31 Explanation:
When first time stkFunc (1,10) will be called then, inside Switch(opcode) the control will go to CaseI, where size=10.
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.
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 32 
12  
120400  
1204  
1034 
Question 32 Explanation:
p = s1+2;
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.
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 34 
140  
150  
160  
170 
Question 34 Explanation:
**ptr = 40
∴ printf (“%d%d”, p + r – p, p + r) will print 140.
Question 35 
10  
11  
12  
13 
Question 35 Explanation:
j = 2*3 / 4+2.0 / 5+8 / 5;
= 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
= 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 36 
230  
240  
250  
260 
Question 36 Explanation:
x = x + f1( ) + f2( ) + f3( ) + f4( )
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
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 37 
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. 
Question 37 Explanation:
int i; // Initially i takes the Garbage value
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.
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 38 
9  
10  
11  
12 
Question 38 Explanation:
func (435)
count = 0 Shift right of 1, which means the number gets half.
while (num)
{ Shift left of 1, which means, the number gets doubled.
count++;
num>>=1;
}
return (count); 435/2 = 217/2 = 108/2 = 54/2 = 27/2 = 13/2 = 6/2 = 3/2 = 1/2 = 0
Count: 1, 2, 3, 4, 5, 6, 7, 8, 9. Χ
(or)
(435)_{10} = (110110011)_{2}
The given program counts total number of bits in binary representation and fails when while (num) becomes all zeroes.
count = 0 Shift right of 1, which means the number gets half.
while (num)
{ Shift left of 1, which means, the number gets doubled.
count++;
num>>=1;
}
return (count); 435/2 = 217/2 = 108/2 = 54/2 = 27/2 = 13/2 = 6/2 = 3/2 = 1/2 = 0
Count: 1, 2, 3, 4, 5, 6, 7, 8, 9. Χ
(or)
(435)_{10} = (110110011)_{2}
The given program counts total number of bits in binary representation and fails when while (num) becomes all zeroes.
Question 39 
Suppose n and p are unsigned int variables in a C program. We wish to set p to ^{n}C_{3}. If n is large, which one of the following statements is most likely to set p correctly?
p = n * (n1) * (n2) / 6;  
p = n * (n1) / 2 * (n2) / 3;  
p = n * (n1) / 3 * (n2) / 2;  
p = n * (n1) * (n2) / 6.0; 
Question 39 Explanation:
n & p are unsigned int variable.
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.
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 40 
1.73  
1.74  
1.75  
1.76 
Question 40 Explanation:
double f(double x){
if( abs(x*x – 3) < 0.01) return x;
else return f(x/2 + 1.5/x);
}
We know that for
1) abs(x*x–3) < 0.01
x^{2} – 3 < 0.01 & – (x^{2} – 3) < 0.01
x^{2} < 3.01 & –x^{2} < 2.99
x < 1.73 & x < 1.72
2) x = x/2 + 1.5/x
x=(x^{2}+(1.5)2)/2x
x=(x^{2}+3)/2x⇒2x^{2}x^{2}=3
x^{2}=3
x=√3=1.732
Whichever the case we get around 1.72, 1.73 & max 1.74.
if( abs(x*x – 3) < 0.01) return x;
else return f(x/2 + 1.5/x);
}
We know that for
1) abs(x*x–3) < 0.01
x^{2} – 3 < 0.01 & – (x^{2} – 3) < 0.01
x^{2} < 3.01 & –x^{2} < 2.99
x < 1.73 & x < 1.72
2) x = x/2 + 1.5/x
x=(x^{2}+(1.5)2)/2x
x=(x^{2}+3)/2x⇒2x^{2}x^{2}=3
x^{2}=3
x=√3=1.732
Whichever the case we get around 1.72, 1.73 & max 1.74.
Question 41 
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. 
Question 41 Explanation:
int f(int j) // if j=50,
{ i is also 50,
static int i = 50; if (i == j) // True
int k; prints something
if (i == j) and called f(50) again,
{ there is no base condition
printf(“something”); to get out of recursion.
k = f(i); So it goes into infinite loop
return 0; when j=50.
}
else return 0;
}
{ i is also 50,
static int i = 50; if (i == j) // True
int k; prints something
if (i == j) and called f(50) again,
{ there is no base condition
printf(“something”); to get out of recursion.
k = f(i); So it goes into infinite loop
return 0; when j=50.
}
else return 0;
}
Question 42 
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 
Question 42 Explanation:
Let be a small matrix.
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.
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 43 
Θ(n^{2})  
Θ(n^{2} log n)  
Θ(n^{3})  
Θ(n^{3} logn) 
Question 43 Explanation:
Outer loop runs for (n/2) times and inner loop runs for (logn) times.
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)
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 44 
None  
2 only  
3 and 4 only  
4 only 
Question 44 Explanation:
The test cases 3 & 4 captures the flaw. The code does not works fine when an old character is replaced by a new character & new character is again replaced by another new character.
1, 2 works fine, 3, 4 carries flaw.
1, 2 works fine, 3, 4 carries flaw.
Question 45 
(i) and (iii)  
(i) and (iv)  
(ii) and (iii)  
(ii) and (iv) 
Question 45 Explanation:
The function can be terminated for all the values which can have factor of 2{(2x)2 == 0}, so (i) is false and (ii) is true.
→ 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.
→ 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 46 
jungle, n, 8, ncestor  
etter, u, 6, ungle  
cetter, k, 6, jungle  
etter, u, 8, ncestor 
Question 46 Explanation:
Lets take the part of program,
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".
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 47 
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 
Question 47 Explanation:
First of all, the swap function just swaps the pointers inside the function and has no effect on the variable being passed.
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
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 48 
8, 12, 7, 23, 8  
8, 8, 7, 23, 7  
12, 12, 27, 31, 23  
12, 12, 27, 31, 56 
Question 48 Explanation:
1) a[0][2] = *(*(a+0)+2)
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.
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 49 
(n == 0)  (m == 1)  
(n == 0) && (m == 1)  
(n == 0)  (m == n)  
(n == 0) && (m == n) 
Question 49 Explanation:
We know that,
^{m}C_{0} = 1
^{n}C_{n} = 1
^{m}C_{0} = 1
^{n}C_{n} = 1
Question 50 
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++]]– 
Question 50 Explanation:
A: Increments the count by 1 at each index that is equal to the ASCII value of the alphabet, it is pointing at.
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.
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 51 
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 51 Explanation:
It is nothing but a pairwise swapping of the linked list.
Question 52 
a[j] – a[i] > S  
a[j] – a[i] < S  
a[i] – a[j] < S  
a[i] – a[j] > S

Question 52 Explanation:
For some 'i' if we find that difference of (A[j]  A[i] < S) we increment 'j' to make this difference wider so that it becomes equal to 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'.
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 53 
5  
8  
9  
20 
Question 53 Explanation:
Hence, 9 is the answer.
Question 54 
x = 1 + x;  
x = 1  x;  
x = x  1;  
x = 1 % x; 
Question 54 Explanation:
x = 1  x
For x = 0, it gives 1.
For x = 1, it gives 0.
For x = 0, it gives 1.
For x = 1, it gives 0.
Question 55 
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 
Question 55 Explanation:
A) push 'a' and pop 'a', push 'b' and pop 'b', push 'c' and pop 'c', and finally push 'd' and pop 'd'. Sequence of popped elements will come to abcd.
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.
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 56 
Question 56 Explanation:
To compute transpose 'j' needs to be started with 'i'. So, A and B are wrong.
In (D) , given statements is wrong as temporary variable needs to be assigned some value and not viceversa.
In (D) , given statements is wrong as temporary variable needs to be assigned some value and not viceversa.
Question 57 
43 80  
42 74  
33 37  
32 32 
Question 57 Explanation:
In first iteration:
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.
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 58 
?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); 
Question 58 Explanation:
getchar( ) = reads a single character at a time from the stdin.
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.
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 59 
AB UV VW VW  
AB UV AB VW  
AB UV UV VW  
AB UV VW UV 
Question 59 Explanation:
→ First print AB.
→ 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).
→ 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 60 
89  
90  
91  
92 
Question 60 Explanation:
Value returned by
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
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 61 
5  
25  
36  
42 
Question 61 Explanation:
If it is call by reference then answer is 42.
If it is call by value then answer is 36.
If it is call by value then answer is 36.
Question 62 
2  
Run time error  
None of the above 
Question 62 Explanation:
Since nothing is said in question. So we will assume by default call by value.
X in the procedure FIND is a local variable. No change will be reflected in global variable X.
X in the procedure FIND is a local variable. No change will be reflected in global variable X.
Question 63 
An unrestricted use of the “goto” statement is harmful because
it makes it more difficult to verify programs  
it increases the running time of the programs  
it increases the memory required for the programs
 
it results in the compiler generating longer machine code 
Question 63 Explanation:
If we use "goto" statements then it leads to structural decomposition of code then it is difficult to verify the programs.
Question 64 
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 64 Explanation:
0, because global m is not modified, m is just passed to formal argument of P.
Question 65 
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 65 Explanation:
0, because n is just passed to formal parameters of Q and no modification in global n.
Question 66 
PARAM, P, Q  
PARAM, P  
PARAM, Q  
P, Q  
none of the above 
Question 66 Explanation:
Since m is defined global it is visible inside all the procedures.
Question 67 
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 
Question 67 Explanation:
Exchanges a and b.
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
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 68 
10  
11  
3  
None of the above 
Question 68 Explanation:
n=3
W(n)=W(3)
Procedure W(var x; int)
begin
x = x+1 = 3+1 = 4
Print x → Print x=4
end
W(n)=W(3)
Procedure W(var x; int)
begin
x = x+1 = 3+1 = 4
Print x → Print x=4
end
Question 69 
An unrestricted use of the "go to" statement is harmful because of which of the following reason (s):
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 69 Explanation:
Dijkstra's argued that unrestricted goto statements should abolished from the higherlevel languages because they complicated the task of analyzing and verifying the correctness of programs.
Question 70 
3, 6  
6, 7  
3, 7  
None of the above. 
Question 70 Explanation:
First procedure Q is called from the main procedure. Q has local variables x and y with values 3 and 4 respectively. This local variable y (value 4) is being parsed to procedure P during call, and received in local variable n inside procedure P. Now as P does not have any local definition for variable x, it will assign the evaluated value of (n+2)/(n3), i.e., (4+2)/(43)=6 to the global variable x, which was previously 7. After the call of procedure P, procedure Q writes the value of local variable x which is still 3. Lastly, the main procedure writes the value of global variable x, which has been changed to 6 inside procedure P. So, the output will be 3, 6.
Question 71 
3, 6  
6, 7  
3, 7  
None of the above 
Question 71 Explanation:
The same sequence of statements will be executed using dynamic scoping. However, as there is no local definition of variable x in procedure P, it will consider the recent definition in the calling sequence, as P is being called from procedure Q, definition of x from Q will be changed to 6 from 3. Now, when Q writes local variables x, 6 will be printed. The write global variable x from main procedure will print 7 (as value of the global variable x has not been changed). So, the output will be 6, 7.
There are 71 questions to complete.