C PROGRAMMING QUESTIONS AND ANSWERS

1. What will be output if you will compile and execute the following c code?
#include<stdio.h>
int main(){
int i=320;
char *ptr=(char *)&i;
printf("%d",*ptr); return 0;
}
(A) 320
(B) 1
(C) 64
(D) Compiler error
(E) None of above
Explanation:
As we know size of int data type is two byte while char pointer can pointer one byte at time.
Memory representation of int i=320
So char pointer ptr is pointing to only first byte as shown above figure.
*ptr i.e. content of first byte is 01000000 and its decimal value is 64.
Answer: (C)

2. What will be output if you will compile and execute the following c code?
#include<stdio.h>
#define x 5+2
int main(){
int i;
i=x*x*x;
printf("%d",i); return 0;
}
(A) 343
(B) 27
(C) 133
(D) Compiler error
(E) None of above
Explanation:
As we know #define is token pasting preprocessor it only paste the value of micro constant in the program before the actual compilation start. If you will see intermediate file you will find:
test.c 1:
test.c 2: void main(){
test.c 3: int i;
test.c 4: i=5+2*5+2*5+2;
test.c 5: printf("%d",i);
test.c 6: }
test.c 7:
You can absorb #define only pastes the 5+2 in place of x in program. So,
i=5+2*5+2*5+2
=5+10+10+2
=27
Answer: (B)

3. What will be output if you will compile and execute the following c code?
#include<stdio.h>
int main(){
char c=125;
c=c+10;
printf("%d",c); return 0;
}
(A) 135
(B) +INF
(C) -121
(D) -8
(E) Compiler error
Explanation:
As we know char data type shows cyclic properties i.e. if you will increase or decrease the char variables beyond its maximum or minimum value respectively it will repeat same value according to following cyclic order:
So,
125+1= 126
125+2= 127
125+3=-128
125+4=-127
125+5=-126
125+6=-125
125+7=-124
125+8=-123
125+9=-122
125+10=-121
Answer: (C)

4. What will be output if you will compile and execute the following c code?
#include<stdio.h>
int main(){
float a=5.2;
if(a==5.2)
printf("Equal");
else if(a<5.2)
printf("Less than");
else
printf("Greater than"); return 0;
}
(A) Equal
(B) Less than
(C) Greater than
(D) Compiler error
(E) None of above
Explanation:
5.2 is double constant in c. In c size of double data is 8 byte while a is float variable. Size of float variable is 4 byte.
So double constant 5.2 is stored in memory as:
101.00 11001100 11001100 11001100 11001100 11001100 11001101
Content of variable a will store in the memory as:
101.00110 01100110 01100110
It is clear variable a is less than double constant 5.2
Since 5.2 is recurring float number so it different for float and double. Number likes 4.5, 3.25, 5.0 will store same values in float and double data type.
Answer: (B)

5. What will be output if you will compile and execute the following c code?
#include<stdio.h>
int main(){
int i=4,x;
x=++i + ++i + ++i;
printf("%d",x); return 0;
}
(A) 21
(B) 18
(C) 12
(D) Compiler error
(E) None of above
Explanation:
In ++a, ++ is pre increment operator. In any mathematical expression pre increment operator first increment the variable up to break point then starts assigning the final value to all variable.
Step 1: Increment the variable I up to break point.
Step 2: Start assigning final value 7 to all variable i in the expression.
So, i=7+7+7=21
Answer: (A)

6. What will be output if you will compile and execute the following c code?
#include<stdio.h>
int main(){
int a=2;
if(a==2){
a=~a+2<<1;
printf("%d",a);
}
else{
break;
} return 0;
}
(A) It will print nothing.
(B) -3
(C) -2
(D) 1
(E) Compiler error
Explanation:
Keyword break is not part of if-else statement. Hence it will show compiler error: Misplaced break
Answer: (E)

7. What will be output if you will compile and execute the following c code?
#include<stdio.h>
int main(){
int a=10;
printf("%d %d %d",a,a++,++a); return 0;
}
(A) 12 11 11
(B) 12 10 10
(C) 11 11 12
(D) 10 10 12
(E) Compiler error
Explanation:
In c printf function follows cdecl parameter passing scheme. In this scheme parameter is passed from right to left direction.
So first ++a will pass and value of variable will be a=10 then a++ will pass now value variable will be a=10 and at the end a will pass and value of a will be a=12.
Answer: (A)

8. What will be output if you will compile and execute the following c code?
#include<stdio.h>
int main(){
char *str="Hello world";
printf("%d",printf("%s",str)); return 0;
}
(A) 11Hello world
(B) 10Hello world
(C) Hello world10
(D) Hello world11
(E) Compiler error
Explanation:
Return type of printf function is integer and value of this integer is exactly equal to number of character including white space printf function prints. So, printf(“Hello world”) will return 13.
Answer: (D)

9. What will be output if you will compile and execute the following c code?
#include <stdio.h>
#include <string.h>
int main(){
char *str=NULL;
strcpy(str,"cquestionbank");
printf("%s",str); return 0;
}
(A) cquestionbank
(B) cquestionbank\0
(C) (null)
(D) It will print nothing
(E) Compiler error
Explanation:
We cannot copy anything using strcpy function to the character pointer pointing to NULL.
Answer: (C)

10. What will be output if you will compile and execute the following c code?
#include <stdio.h>
#include <string.h>
int main(){
int i=0;
for(;i<=2;)
printf(" %d",++i); return 0;
}
(A) 0 1 2
(B) 0 1 2 3
(C) 1 2 3
(D) Compiler error
(E) Infinite loop
Explanation:
In for loop each part is optional.
Answer: (C)

11. What will be output if you will compile and execute the following c code?
#include<stdio.h>
int main(){
int x;
for(x=1;x<=5;x++);
printf("%d",x); return 0;
}
(A) 4
(B) 5
(C) 6
(D) Compiler error
(E) None of above
Explanation:
Body of for loop is optional. In this question for loop will execute until value of variable x became six and condition became false.
Answer: (C)

12. What will be output if you will compile and execute the following c code?
#include<stdio.h>
int main(){
printf("%d",sizeof(5.2)); return 0;
}
(A) 2
(B) 4
(C) 8
(D) 10
(E) Compiler error
Explanation:
Default type of floating point constant is double. So 5.2 is double constant and its size is 8 byte.
Answer: (C)

13. What will be output if you will compile and execute the following c code?
#include <stdio.h>
#include <string.h>
int main(){
char c='\08';
printf("%d",c); return 0;
}
(A) 8
(B) ‟8‟
(C) 9
(D) null
(E) Compiler error
Explanation:
In c any character is starting with character „\‟ represents octal number in character. As we know octal digits are: 0, 1, 2, 3, 4, 5, 6, and 7. So 8 is not an octal digit. Hence „\08‟ is invalid octal character constant.
Answer: (E)

14. What will be output if you will compile and execute the following c code?
#include<stdio.h>
#define call(x,y) x##y
int main(){
int x=5,y=10,xy=20;
printf("%d",xy+call(x,y)); return 0;
}
(A) 35
(B) 510
(C) 15
(D) 40
(E) None of above
Explanation:
## is concatenation c preprocessor operator. It only concatenates the operands i.e.
a##b=ab
If you will see intermediate file then you will find code has converted into following intermediate code before the start of actual compilation.
Intermediate file:
test.c 1:
test.c 2: void main(){
test.c 3: int x=5,y=10,xy=20;
test.c 4: printf("%d",xy+xy);
test.c 5: }
test.c 6:
It is clear call(x, y) has replaced by xy.
Answer: (D)

15. What will be output if you will compile and execute the following c code? #include<stdio.h>
int * call();
int main(){
int *ptr;
ptr=call();
printf("%d",*ptr); return 0;
}
int * call(){
int a=25;
a++;
return &a;
}
(A) 25
(B) 26
(C) Any address
(D) Garbage value
(E) Compiler error
Explanation:
In this question variable a is a local variable and its scope and visibility is within the function call. After returning the address of a by function call variable a became dead while pointer ptr is still pointing to address of variable a. This problem is known as dangling pointer problem.
Answer: (D)

16. What is error in following declaration?
struct outer{
int a;
struct inner{
char c;
};
};
(A) Nesting of structure is not allowed in c.
(B) It is necessary to initialize the member variable.
(C) Inner structure must have name.
(D) Outer structure must have name.
(E) There is not any error.
Explanation:
It is necessary to assign name of inner structure at the time of declaration otherwise we cannot access the member of inner structure. So correct declaration is:
struct outer{
int a;
struct inner{
char c;
}name;
};
Answer: (C)

17. What will be output if you will compile and execute the following c code?
#include<stdio.h>
int main(){
int array[]={10,20,30,40};
printf("%d",-2[array]); return 0;
}
(A) -60
(B) -30
(C) 60
(D) Garbage value
(E) Compiler error
Explanation:
In c,
array[2]=*(array+2)=*(2+array)=2[array]=30
Answer: (B)

18. What will be output if you will compile and execute the following c code?
#include<stdio.h>
int main(){
int i=10;
static int x=i;
if(x==i)
printf("Equal");
else if(x>i)
printf("Greater than");
else
printf("Less than");
return 0;
}
(A) Equal
(B) Greater than
(C) Less than
(D) Compiler error
(E) None of above
Explanation:
Static variables are load time entity while auto variables are run time entity. We cannot initialize any load time variable by the run time variable.
In this example i is run time variable while x is load time variable.
Answer: (D)

19. What will be output if you will compile and execute the following c code?
#include<stdio.h>
#define max 5;
int main(){
int i=0;
i=max++;
printf("%d",i++); return 0;
}
(A) 5
(B) 6
(C) 7
(D) 0
(E) Compiler error
Explanation:
#define is token pasting preprocessor. If you will see intermediate file: test.i
test.c 1:
test.c 2: void main(){
test.c 3: int i=0;
test.c 4: i=5++;
test.c 5: printf("%d",i++);
test.c 6: }
test.c 7:
It is clear macro constant max has replaced by 5. It is illegal to increment the constant number. Hence compiler will show Lvalue required.
Answer: (D)

20. What will be output if you will compile and execute the following c code? #include<stdio.h>
int main(){
double far* p,q;
printf("%d",sizeof(p)+sizeof q);
return 0;
}
(A) 12
(B) 8
(C) 4
(D) 1
(E) Compiler error
Explanation:
It is clear p is far pointer and size of far pointer is 4 byte while q is double variable and size of double variable is 8 byte.
Answer: (A)