C programming online test


C programming online test questions answers and explanation for freshers



Topic:
Total questions:
Total marks:
Correct answer:
Total time:
Incorrect answer:
Passing Marks:
1
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    int i;
    for(i=0;i<5;i++){
         int i=10;
         printf(" %d",i);
         i++;
    }
    return 0;
}
(A)
10 11 12 13 14
(B)
10 10 10 10 10
(C)
0 1 2 3 4
(D)
Compilation error
                         
Explanation:
Default storage class of local variable is auto. Scope of auto variables are block in which it has been declared. When program control goes out of the scope auto variables are dead. So variable i which has been declared inside for loop has scope within loop and in each iteration variable i is dead and re-initialized.
Note: If we have declared two variables of same name but different scope then local variable will have higher priority.
2
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    register a,b,x;
    scanf("%d %d",&a,&b);
    x=a+~b;
    printf("%d",x);
    return 0;
}
(A)
0
(B)
It will be difference of a and b
(C)
It will be addition of a and b
(D)
Compilation error
                         
Explanation:
Register variables are stored in CPU. So it has not memory address. Hence it is incorrect to write &a.
3
What will be output if you will execute following c code?

#include<stdio.h>
auto int a=5;
int main(){
    int x;
    x=~a+a&a+a<<a;
    printf("%d",x);
    return 0;
}
(A)
5
(B)
0
(C)
153
(D)
Compilation error
                         
Explanation:
We cannot declare auto variable outside of any function since it auto variables gets are created (i.e. gets memory) at run time.
4
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    register int a,b;
    int c;
    scanf("%d%d",&a,&b);
    c=~a + ~b + ++a + b++;
    printf(" %d",c);
    return 0;
}
//User input is: 1 2
(A)
-1
(B)
0
(C)
1
(D)
Compilation error
                         
Explanation:
Register variables are stored in CPU. So it has not memory address. Hence it is incorrect to write &a.
5
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    int arr[3]={10,20,30};
    int x=0;
    x = ++arr[++x] + ++x + arr[--x];
    printf("%d ",x);
    return 0;
}
(A)
22
(B)
23
(C)
43
(D)
44
                         
Explanation:
In Turbo C 3.0 and 4.5 compilers
Output: 43

Consider on expression:

= ++arr[++x] + ++x + arr[--x] //x = 0 + 1
= ++arr[++x] + ++x + arr[--x] //x = 1 + 1
= ++arr[++x] + ++x + arr[--x] //x = 2 - 1
= ++arr[1] + 1 + arr[1] //x = 1
= ++arr[1] + 1 + arr[1]  //arr[1] = 20+1
= arr[1] + 1 + arr[1] //arr[1] = 21
= 21 + 1 + 21
= 43
In Linux GCC complier
Output: 44

Consider on expression:
= ++arr[++x] + ++x + arr[--x] //x = 0 + 1
= ++arr[1] + ++x + arr[--x] ////x = 1 + 1
= ++arr[++x] + 2 + arr[--x] //x = 2 - 1
= ++arr[1] + 2 + arr[1] //arr[1] = 20+1
= arr[1] + 1 + arr[1] //arr[1] = 21
= 21 + 2 + 21
= 44
6
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    int a[]={10,20,30,40};
    int i=3,x;
    x=1*a[--i]+2*a[--i]+3*a[--i];
    printf("%d",x);
    return 0;
}
(A)
30
(B)
60
(C)
90
(D)
Compilation error
                         
Explanation:
In Turbo C 3.0 and 4.5 compilers
Output: 60

Consider on expression:
= 1 * a[--i] + 2 * a[--i] + 3 * a[--i] //i = 3 - 2
= 1 * a[--i] + 2 * a[--i] + 3 * a[--i] //i = 2 - 1
= 1 * a[--i] + 2 * a[--i] + 3 * a[--i] //i = 1 - 1
= 1 * a[0] + 2 * a[0] + 3 * a[0] //i = 0
= 1 * 10 + 2 * 10 + 3 * 10 //a[0] = 10
= 10 + 20 + 30
= 60

In Linux GCC complier
Output: 90

Consider on expression:
= 1 * a[--i] + 2 * a[--i] + 3 * a[--i] //i = 3 - 2
= 1 * a[--i] + 2 * a[--i] + 3 * a[--i] //i = 2 - 1
= 1 * a[1] + 2 * a[1] + 3 * a[--i] //i = 1 - 1
= 1 * a[1] + 2 * a[1] + 3 * a[0]
= 1 * 20 + 2 * 20 + 3 * 10
= 20 + 40 + 30
= 90
7
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    static int a[][2][3]={0,1,2,3,4,5,6,7,8,9,10,11,12};
    int i=-1;
    int d;
    d=a[i++][++i][++i];
    printf("%d",d);
    return 0;
}
(A)
9
(B)
10
(C)
11
(D)
Compilation error
                         
Explanation:
= a[i++][++i][++i] //i = -1 + 1
= a[i++][++i][++i] //i = 0 + 1
= a[1][1][1] //i = 1 + 1
= 10
8
What will be output if you will execute following c code?

#include<stdio.h>
int f(int);
int main(){
    int i=3,val;
    val=sizeof (f(i)+ +f(i=1)+ +f(i-1));
    printf("%d %d",val,i);  
    return 0; 
}
int f(int num){
        return num*5;
}
(A)
2 3
(B)
4 3
(C)
3 2
(D)
Compilation error
                         
Explanation:
Turbo C 3.0 and Turbo C 4.5 compiler:
2 3
Linux GCC complier:
4 3

Any expression inside sizeof operator is never changed the value of the any variable. So value of variable i will remain 3. After the evaluation of expression inside sizeof operator we will get an integer value. So value of variable val will be sizeof int data type.

Note: Size of into in turbo C 3.0 and 4.5 is two byte while Linux gcc complier is four byte
9
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    int x,a=3;
    x=+ +a+ + +a+ + +5;
    printf("%d  %d",x,a);
    return 0;
}
(A)
10 3
(B)
11 3
(C)
10 5
(D)
Compilation error
                         
Explanation:
Consider on expression: + +a
Here both + are unary plus operation. So
= + +a+ + +a+ + +5;
= + +3+ + +3+ + 5
= 3+ 3+ 5
= 11

Note: Increment operator ++ cannot have space between two plus symbol.
10
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    int num,i=0;
    num=-++i+ ++-i;
    printf("%d",num);
    return 0;
}
(A)
0
(B)
1
(C)
-2
(D)
Compilation error
                         
Explanation:
After operation of any operator on operand it returns constant value. Here we are performing unary minus operator on variable i so it will return a constant value and we can perform ++ operation on constant.  
11
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    int num,a=5;
    num=-a--+ +++a;
    printf("%d  %d",num,a);
    return 0;
}
(A)
1 5
(B)
-1 6
(C)
1 6
(D)
0 5
                         
Explanation:
= -a--+ +++a
= -a-- + + ++a
= -a-- + + ++a
= -6 + + 6 //a = 6 -1
= -6 + 6 //a = 5
= 0
12
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    int num,a=15;
    num=- - - -a--;
    printf("%d  %d",num,a);
    return 0;
}
(A)
15 14
(B)
14 15
(C)
14 14
(D)
15 15
                         
Explanation:
= - - - -a
= - - - -15 //a = 15 – 1
= 15  //a = 14
13
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    int x,a=2;
    x=++a,++a,a++;
    printf("%d  %d",x,a);
    return 0;
}
(A)
5 5
(B)
3 5
(C)
4 5
(D)
5 4
                         
Explanation:
x = ++a, ++a, a++
x = 3, ++a, a++ // a = 2 + 1
x = 3, ++a, a++ // = operator has higher precedence than comma operator
x = 3, ++a, a++ // a = 3 + 1
x = 3, 4, a++ 
x = 3, 4, 4 // a = 4 + 1
x = 3 // a = 5
14
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    int x,i=2;
    x=~-!++i;
    printf("%d",x);
    return 0;
}
(A)
-2
(B)
-1
(C)
0
(D)
1
                         
Explanation:
= ~-!++i
= ~-!3 //i = 2 + 1
= ~-0 //!3 = 0
= ~0 //-0 = 0
= -(0 + 1) //~ is 1's complement operator.
= -1 
15
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    static double *p,*q,*r,*s,t=5.0;
    double **arr[]={&p,&q,&r,&s};
    int i;
    *p=*q=*r=*s=t;
    for(i=0;i<4;i++)
        printf("%.0f  ",**arr[i]);
    return 0;
}
(A)
5 5 5 5 5 
(B)
5 6 7 8 9
(C)
Infinite loop
(D)
Run time error

                         
Explanation:
Turbo C 3.0:
5 5 5 5 5 
Turbo C 4.5 and Linux GCC complier:
Run time error
16
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    float x;
    x=0.35==3.5/10;
    printf("%f",x);
    return 0;
}
(A)
0.000000
(B)
1.000000
(C)
0.350000
(D)
Compilation error
                         
Explanation:
Turbo C 3.0 and Turbo C 4.5 compiler:
Output: 0.000000
3.5/10 is little greater than .35

Linux GCC compilers:
Output: 1.000000

Note: == is logic operator. It returns 1 if both operands are equal otherwise it returns 0.
17
#include<stdio.h>
int main(){
    int arr[]={6,12,18,24};
    int x=0;
    x=arr[1]+(arr[1]=2);
    printf("%d",x);
    return 0;
}
(A)
4
(B)
8
(C)
14
(D)
Compilation error
                         
Explanation:
= arr[1] + (arr[1] = 2) //arr[i] = 2
= arr[1] + arr[1]
= 2 + 2
= 4
18
What will be output if you will execute following c code?

#include<stdio.h>
int sq(int);
int main(){
    int a=1,x;
    x=sq(++a)+sq(a++)+sq(a++);
    printf("%d",x);
return 0;
}
int sq(int num){
    return num*num;
}
(A)
15
(B)
16
(C)
17
(D)
18
                         
Explanation:
= sq(++a) + sq(a++) + sq(a++) //a= 1 + 1
= sq(2) + sq(2) + sq(a++) //a = 2 + 1
= sq(2) + sq(2) + sq(3//a = 3 + 1
= 4 + 4 + 9
= 17


Note: Pre-increment fist increment then assign while post increment operator first assign then increment.
19
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    printf("%c",*"abcde");
return 0;
}
(A)
acbcd
(B)
e
(C)
a
(D)
NULL
                         
Explanation:
String constant "abcde" will return memory address of first character of the string constant. *"abcde" will return the first character of string constant.  
20
What will be output if you will execute following c code?

#include<stdio.h>
int main(){
    printf("%d","abcde"-"abcde");
return 0;
}
(A)
0
(B)
-1
(C)
1
(D)
Garbage
                         
Explanation:
Memory address of string constants depends upon operating system.
Total correct answers:
Total incorrect answers:
Total not attempted questions:
Marks obtained:
Result:

It is C programming language online test or quiz objective types questions answers with explanation for interview or written test. If you have any queries or suggestions, please share it.

12 comments:

  1. which one is standard compiler? turbo c dev c++ or lunix?

    ReplyDelete
  2. can anybody explain this???


    #include

    main()
    {
    int **p,q;
    p = (int **)5;
    q = 10;
    printf("%d",q+p);
    return 0;
    }

    ReplyDelete
  3. i am a first year INFORMATION TECHNOLOGY student.at UE .this really help me to understand my course more..thank you :)

    ReplyDelete
  4. arr[ ]={4,5,8,8,4};
    x=arr[1]+(arr[1]=2);
    while executing this line which one should be executed first whether arr[ 1 ] or (arr[ 1 ]=2) helpme

    ReplyDelete
  5. In q6 the gcc compilers output is 110 . pls verify and tell us the reason

    ReplyDelete
  6. will you provide certificate for this test if we pass

    ReplyDelete
  7. #include
    int main(){
    int a[]={10,20,30,40};
    int i=3,x;
    x=1*a[--i]+2*a[--i]+3*a[--i];
    printf("%d",x);
    return 0;
    }


    here the o/p is 100 but you have given the ans as 60 can u explain this for me

    ReplyDelete
  8. I suggest you refactor this test and allow for different compiler treatment of undefined cases and other moot points. FYI, some of your "Compilation errors" actually compile on my machine, int size is not specified by the standard, your float expression evaluation on my machine was actually not done as float and so on...

    ReplyDelete
  9. chutiyo ans is 100 for sixth ques

    ReplyDelete