C Programming Functions Online Test, C Programming Mock Test

C Programming Functions Online Test

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C Programming Functions Online Test. C Programming Question and Answers in English. C Programming Functions Online mock test paper is free for all students and Very Helpful for Exam Preparation. C Programming Functions Online Quiz. C Programming Online Mock test for Functions Topic. Here we are providing C Programming Functions Online Test Series in English. Check C Programming Mock Test Series 2024-2024.

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Question 1 of 30

1. Question

The keyword used to transfer control from a function back to the calling function is

switch
goto
go back
return

Correct

The keyword return is used to transfer control from a function back to the calling function.

Example:

#include<stdio.h>
int add(int, int); /* Function prototype */

int main()
{
    int a = 4, b = 3, c;
    c = add(a, b);
    printf("c = %d\n", c);
    return 0;
}
int add(int a, int b)
{
/* returns the value and control back to main() function */
   return (a+b);
}

Output:
c = 7

Incorrect

The keyword return is used to transfer control from a function back to the calling function.

Example:

#include<stdio.h>
int add(int, int); /* Function prototype */

int main()
{
    int a = 4, b = 3, c;
    c = add(a, b);
    printf("c = %d\n", c);
    return 0;
}
int add(int a, int b)
{
/* returns the value and control back to main() function */
   return (a+b);
}

Output:
c = 7

Question 2 of 30

2. Question
What is the notation for following functions?
1. int f(int a, float b) { /* Some code */ } 2. int f(a, b) int a; float b; { /* Some code */ }
- 1. KR Notation 2. ANSI Notation
- 1. Pre ANSI C Notation 2. KR Notation
- 1. ANSI Notation 2. KR Notation
- 1. ANSI Notation 2. Pre ANSI Notation
Correct

KR Notation means Kernighan and Ritche Notation.

Incorrect

KR Notation means Kernighan and Ritche Notation.
Question 3 of 30

3. Question
How many times the program will print “IndiaBIX” ?
#include<stdio.h> int main() { printf("IndiaBIX"); main(); return 0; }
- Infinite times
- 32767 times
- 65535 times
- Till stack overflows
Correct

A call stack or function stack is used for several related purposes, but the main reason for having one is to keep track of the point to which each active subroutine should return control when it finishes executing.

A stack overflow occurs when too much memory is used on the call stack.

Here function main() is called repeatedly and its return address is stored in the stack. After stack memory is full. It shows stack overflow error.

Incorrect

A call stack or function stack is used for several related purposes, but the main reason for having one is to keep track of the point to which each active subroutine should return control when it finishes executing.

A stack overflow occurs when too much memory is used on the call stack.

Here function main() is called repeatedly and its return address is stored in the stack. After stack memory is full. It shows stack overflow error.
Question 4 of 30

4. Question
What will be the output of the program in 16 bit platform (Turbo C under DOS)?
#include<stdio.h> int main() { int fun(); int i; i = fun(); printf("%d\n", i); return 0; } int fun() { _AX = 1990; }
- Garbage value
- 0 (Zero)
- 1990
- No output
Correct

Turbo C (Windows): The return value of the function is taken from the Accumulator _AX=1990.

But it may not work as expected in GCC compiler (Linux).

Incorrect

Turbo C (Windows): The return value of the function is taken from the Accumulator _AX=1990.

But it may not work as expected in GCC compiler (Linux).
Question 5 of 30

5. Question
What will be the output of the program?
#include<stdio.h> void fun(int*, int*); int main() { int i=5, j=2; fun(&i, &j); printf("%d, %d", i, j); return 0; } void fun(int *i, int *j) { *i = *i**i; *j = *j**j; }
- 5, 2
- 10, 4
- 2, 5
- 25, 4
Correct

Step 1: int i=5, j=2; Here variable i and j are declared as an integer type and initialized to 5 and 2 respectively.

Step 2: fun(&i, &j); Here the function fun() is called with two parameters &i and &j (The & denotes call by reference. So the address of the variable i and j are passed. )

Step 3: void fun(int *i, int *j) This function is called by reference, so we have to use * before the parameters.

Step 4: *i = *i**i; Here *i denotes the value of the variable i. We are multiplying 5*5 and storing the result 25 in same variable i.

Step 5: *j = *j**j; Here *j denotes the value of the variable j. We are multiplying 2*2 and storing the result 4 in same variable j.

Step 6: Then the function void fun(int *i, int *j) return back the control back to main() function.

Step 7: printf(“%d, %d”, i, j); It prints the value of variable i and j.

Hence the output is 25, 4.

Incorrect

Step 1: int i=5, j=2; Here variable i and j are declared as an integer type and initialized to 5 and 2 respectively.

Step 2: fun(&i, &j); Here the function fun() is called with two parameters &i and &j (The & denotes call by reference. So the address of the variable i and j are passed. )

Step 3: void fun(int *i, int *j) This function is called by reference, so we have to use * before the parameters.

Step 4: *i = *i**i; Here *i denotes the value of the variable i. We are multiplying 5*5 and storing the result 25 in same variable i.

Step 5: *j = *j**j; Here *j denotes the value of the variable j. We are multiplying 2*2 and storing the result 4 in same variable j.

Step 6: Then the function void fun(int *i, int *j) return back the control back to main() function.

Step 7: printf(“%d, %d”, i, j); It prints the value of variable i and j.

Hence the output is 25, 4.
Question 6 of 30

6. Question
What will be the output of the program?
#include<stdio.h> int i; int fun(); int main() { while(i) { fun(); main(); } printf("Hello\n"); return 0; } int fun() { printf("Hi"); }
- Hello
- Hi Hello
- No output
- Infinite loop
Correct

Step 1: int i; The variable i is declared as an integer type.

Step 1: int fun(); This prototype tells the compiler that the function fun() does not accept any arguments and it returns an integer value.

Step 1: while(i) The value of i is not initialized so this while condition is failed. So, it does not execute the while block.

Step 1: printf(“Hello\n”); It prints “Hello”.

Hence the output of the program is “Hello”.

Incorrect

Step 1: int i; The variable i is declared as an integer type.

Step 1: int fun(); This prototype tells the compiler that the function fun() does not accept any arguments and it returns an integer value.

Step 1: while(i) The value of i is not initialized so this while condition is failed. So, it does not execute the while block.

Step 1: printf(“Hello\n”); It prints “Hello”.

Hence the output of the program is “Hello”.

Question 7 of 30

7. Question

What will be the output of the program?

#include<stdio.h>
int reverse(int);

int main()
{
    int no=5;
    reverse(no);
    return 0;
}
int reverse(int no)
{
    if(no == 0)
        return 0;
    else
        printf("%d,", no);
    reverse (no--);
}

Print 5, 4, 3, 2, 1
Print 1, 2, 3, 4, 5
Print 5, 4, 3, 2, 1, 0
Infinite loop

Correct

Step 1: int no=5; The variable no is declared as integer type and initialized to 5.

Step 2: reverse(no); becomes reverse(5); It calls the function reverse() with ‘5’ as parameter.

The function reverse accept an integer number 5 and it returns ‘0’(zero) if(5 == 0) if the given number is ‘0’(zero) or else printf(“%d,”, no); it prints that number 5 and calls the function reverse(5);.

The function runs infinetely because the there is a post-decrement operator is used. It will not decrease the value of ‘n’ before calling the reverse() function. So, it calls reverse(5) infinitely.

Note: If we use pre-decrement operator like reverse(–n), then the output will be 5, 4, 3, 2, 1. Because before calling the function, it decrements the value of ‘n’.

Incorrect

Step 1: int no=5; The variable no is declared as integer type and initialized to 5.

Step 2: reverse(no); becomes reverse(5); It calls the function reverse() with ‘5’ as parameter.

Note: If we use pre-decrement operator like reverse(–n), then the output will be 5, 4, 3, 2, 1. Because before calling the function, it decrements the value of ‘n’.

Question 8 of 30

8. Question

What will be the output of the program?

#include<stdio.h>
void fun(int);
typedef int (*pf) (int, int);
int proc(pf, int, int);

int main()
{
    int a=3;
    fun(a);
    return 0;
}
void fun(int n)
{
    if(n > 0)
    {
        fun(--n);
        printf("%d,", n);
        fun(--n);
    }
}

0, 2, 1, 0,
1, 1, 2, 0,
0, 1, 0, 2,
0, 1, 2, 0,

Correct

Incorrect

Question 9 of 30

9. Question

What will be the output of the program?

#include<stdio.h>
int sumdig(int);
int main()
{
    int a, b;
    a = sumdig(123);
    b = sumdig(123);
    printf("%d, %d\n", a, b);
    return 0;
}
int sumdig(int n)
{
    int s, d;
    if(n!=0)
    {
        d = n%10;
        n = n/10;
        s = d+sumdig(n);
    }
    else
        return 0;
    return s;
}

4, 4
3, 3
6, 6
12, 12

Correct

Incorrect

Question 10 of 30

10. Question

What will be the output of the program?

#include<stdio.h>

int main()
{
    void fun(char*);
    char a[100];
    a[0] = 'A'; a[1] = 'B';
    a[2] = 'C'; a[3] = 'D';
    fun(&a[0]);
    return 0;
}
void fun(char *a)
{
    a++;
    printf("%c", *a);
    a++;
    printf("%c", *a);
}

AB
BC
CD
No output

Correct

Incorrect

Question 11 of 30

11. Question
What will be the output of the program?
#include<stdio.h> int main() { int fun(int); int i = fun(10); printf("%d\n", --i); return 0; } int fun(int i) { return (i++); }
- 9
- 10
- 11
- 8
Correct

Step 1: int fun(int); Here we declare the prototype of the function fun().

Step 2: int i = fun(10); The variable i is declared as an integer type and the result of the fun(10) will be stored in the variable i.

Step 3: int fun(int i){ return (i++); } Inside the fun() we are returning a value return(i++). It returns 10. because i++ is the post-increement operator.

Step 4: Then the control back to the main function and the value 10 is assigned to variable i.

Step 5: printf(“%d\n”, –i); Here –i denoted pre-increement. Hence it prints the value 9.

Incorrect

Step 1: int fun(int); Here we declare the prototype of the function fun().

Step 2: int i = fun(10); The variable i is declared as an integer type and the result of the fun(10) will be stored in the variable i.

Step 3: int fun(int i){ return (i++); } Inside the fun() we are returning a value return(i++). It returns 10. because i++ is the post-increement operator.

Step 4: Then the control back to the main function and the value 10 is assigned to variable i.

Step 5: printf(“%d\n”, –i); Here –i denoted pre-increement. Hence it prints the value 9.

Question 12 of 30

12. Question

What will be the output of the program?

#include<stdio.h>
int check (int, int);

int main()
{
    int c;
    c = check(10, 20);
    printf("c=%d\n", c);
    return 0;
}
int check(int i, int j)
{
    int *p, *q;
    p=&i;
    q=&j;
    i>=45 ? return(*p): return(*q);
}

Print 10
Print 20
Print 1
Compile error

Correct

There is an error in this line i>=45 ? return(*p): return(*q);. We cannot use return keyword in the terenary operators.

Incorrect

There is an error in this line i>=45 ? return(*p): return(*q);. We cannot use return keyword in the terenary operators.

Question 13 of 30

13. Question

What will be the output of the program?

#include<stdio.h>
int fun(int, int);
typedef int (*pf) (int, int);
int proc(pf, int, int);

int main()
{
    printf("%d\n", proc(fun, 6, 6));
    return 0;
}
int fun(int a, int b)
{
   return (a==b);
}
int proc(pf p, int a, int b)
{
   return ((*p)(a, b));
}

6
1
0
-1

Correct

Incorrect

Question 14 of 30

14. Question
What will be the output of the program?
#include<stdio.h> int main() { int i=1; if(!i) printf("IndiaBIX,"); else { i=0; printf("C-Program"); main(); } return 0; }
- prints "IndiaBIX, C-Program" infinitely
- prints "C-Program" infinetly
- prints "C-Program, IndiaBIX" infinitely
- Error: main() should not inside else statement
Correct

Step 1: int i=1; The variable i is declared as an integer type and initialized to 1(one).

Step 2: if(!i) Here the !(NOT) operator reverts the i value 1 to 0. Hence the if(0) condition fails. So it goes to else part.

Step 3: else { i=0; In the else part variable i is assigned to value 0(zero).

Step 4: printf(“C-Program”); It prints the “C-program”.

Step 5: main(); Here we are calling the main() function.

After calling the function, the program repeats from step 1 to step 5 infinitely.

Hence it prints “C-Program” infinitely.

Incorrect

Step 1: int i=1; The variable i is declared as an integer type and initialized to 1(one).

Step 2: if(!i) Here the !(NOT) operator reverts the i value 1 to 0. Hence the if(0) condition fails. So it goes to else part.

Step 3: else { i=0; In the else part variable i is assigned to value 0(zero).

Step 4: printf(“C-Program”); It prints the “C-program”.

Step 5: main(); Here we are calling the main() function.

After calling the function, the program repeats from step 1 to step 5 infinitely.

Hence it prints “C-Program” infinitely.

Question 15 of 30

15. Question

What will be the output of the program?

#include<stdio.h>

int addmult(int ii, int jj)
{
    int kk, ll;
    kk = ii + jj;
    ll = ii * jj;
    return (kk, ll);
}

int main()
{
    int i=3, j=4, k, l;
    k = addmult(i, j);
    l = addmult(i, j);
    printf("%d %d\n", k, l);
    return 0;
}

12 12
No error, No output
Error: Compile error
None of above

Correct

Incorrect

Question 16 of 30

16. Question

What will be the output of the program?

#include<stdio.h>
int i;
int fun1(int);
int fun2(int);

int main()
{
    extern int j;
    int i=3;
    fun1(i);
    printf("%d,", i);
    fun2(i);
    printf("%d", i);
    return 0;
}
int fun1(int j)
{
    printf("%d,", ++j);
    return 0;
}
int fun2(int i)
{
    printf("%d,", ++i);
    return 0;
}
int j=1;

3, 4, 4, 3
4, 3, 4, 3
3, 3, 4, 4
3, 4, 3, 4

Correct

Step 1: int i; The variable i is declared as an global and integer type.

Step 2: int fun1(int); This prototype tells the compiler that the fun1() accepts the one integer parameter and returns the integer value.

Step 3: int fun2(int); This prototype tells the compiler that the fun2() accepts the one integer parameter and returns the integer value.

Step 4: extern int j; Inside the main function, the extern variable j is declared and defined in another source file.

Step 5: int i=3; The local variable i is defines as an integer type and initialized to 3.

Step 6: fun1(i); The fun1(i) increements the given value of variable i prints it. Here fun1(i) becomesfun1(3) hence it prints ‘4’ then the control is given back to the main function.

Step 7: printf(“%d,”, i); It prints the value of local variable i. So, it prints ‘3’.

Step 8: fun2(i); The fun2(i) increements the given value of variable i prints it. Here fun2(i) becomes fun2(3) hence it prints ‘4’ then the control is given back to the main function.

Step 9: printf(“%d,”, i); It prints the value of local variable i. So, it prints ‘3’.

Hence the output is “4 3 4 3”.