Functions Questions and Answers

Q: What is the output of this C code? int *i; int main() { if (i == NULL) printf("true\n"); return 0; }

Correct answer: truei is at file scope with no initialiser, so it is zero-initialised - and for a pointer that means the null pointer. The comparison with NULL therefore succeeds and true is printed.

Q: What is the output of the following program? #include <stdio.h> void swap(int m, int n) { int x = m; m = n; n = x; } main() { int x=5, y=3; swap(x,y); printf("%d %d", x, y); }

Correct answer: 5 3C passes arguments by value, so swap receives copies and exchanges the copies. The originals in main are untouched, and the output is 5 3. To swap the caller’s variables the function must take pointers and be called as swap(&x, &y).

Q: The variable declaration with no storage class specified is by default:

Correct answer: autoWith no storage class specified, a variable declared inside a block is auto. (At file scope the default is different: external linkage and static storage duration.)

Q: Can a function return a structure in C?

Correct answer: trueA function may return a structure, and the whole structure is copied back to the caller. It is the standard way of returning several values at once - and it is why the language does not need multiple return values. Returning a large structure copies a good deal of data, so a pointer is often preferred for big ones.

Q: What is the output of this C code? int x = 5; void main() { int x = 3; printf("%d", x); { x = 4; } printf("%d", x); }

Correct answer: 3 4The inner x = 4; is an assignment, not a declaration, so it changes the x already in scope - main's local. The first printf shows 3 and the second, after the block, shows 4. The file-scope x = 5 is shadowed throughout and never seen.

Q: What is the output of this C code? void main() { m(); m(); } void m() { static int x = 5; x++; printf("%d", x); }

Correct answer: 6 7A static local is initialised once, before the program starts, and keeps its value between calls - its storage lasts as long as the program even though its name is local. So the first call prints 6 and the second 7. An ordinary local would print 6 both times.

Q: The output of the code below is void m(int k) { printf("hi"); } void m(double k) { printf("hello"); } void main() { m(3); }

Correct answer: Compile time errorC has no function overloading: two functions cannot share a name, whatever their parameters. The compiler reports a redefinition. In C++ this would compile, and m(3) would select the int version.

61.

What is the scope of an external variable?

A.Whole source file in which it is defined
B.From the point of declaration to the end of the file in which it is defined
C.Any source file in a program
D.From the point of declaration to the end of the file being compiled

Answer: D

Correct answer: From the point of declaration to the end of the file being compiled

Scope runs from the point of declaration to the end of the file - a variable declared at the bottom of a file is not visible to functions above it. Being reachable from other files is a matter of linkage, which is a different property; an extern declaration in the other file is what brings it into scope there.

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62.

What is the scope of a function?

A.Whole source file in which it is defined
B.From the point of declaration to the end of the file in which it is defined
C.Any source file in a program
D.From the point of declaration to the end of the file being compiled

Answer: D

Correct answer: From the point of declaration to the end of the file being compiled

A function’s name is in scope from its declaration to the end of the file - which is exactly why prototypes are put at the top or in a header. Its linkage is external, so other files can call it once they have declared it.

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63.

What is the output of this C code?

int *i;
    int main()
    {
        if (i == NULL)
            printf("true\n");
        return 0;
    }

A.true
B.true only if NULL value is 0
C.Compile time error
D.Nothing

Answer: A

Correct answer: true

i is at file scope with no initialiser, so it is zero-initialised - and for a pointer that means the null pointer. The comparison with NULL therefore succeeds and true is printed.

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64.

What is the output of the following program?

#include <stdio.h>

void swap(int m, int n)
{
   int x = m;
   
   m = n;
   n = x;
}
main()
{
   int x=5, y=3;

   swap(x,y);
   printf("%d %d", x, y);
}

A.3 5
B.5 3
C.5 5
D.Compile error

Answer: B

Correct answer: 5 3

C passes arguments by value, so swap receives copies and exchanges the copies. The originals in main are untouched, and the output is 5 3. To swap the caller’s variables the function must take pointers and be called as swap(&x, &y).

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65.

The variable declaration with no storage class specified is by default:

A.auto
B.extern
C.static
D.register

Answer: A

Correct answer: auto

With no storage class specified, a variable declared inside a block is auto. (At file scope the default is different: external linkage and static storage duration.)

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66.

Can a function return a structure in C?

A.true
B.false
C.Depends on the compiler
D.Depends on the standard

Answer: A

Correct answer: true

A function may return a structure, and the whole structure is copied back to the caller. It is the standard way of returning several values at once - and it is why the language does not need multiple return values. Returning a large structure copies a good deal of data, so a pointer is often preferred for big ones.

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67.

Is it true that too many recursive calls may result into stack overflow?

A.Yes
B.No

Answer: A

Yes, too many recursive calls may result into stack overflow. because when a function is called its return address is stored in stack.

After sometime the stack memory will be filled completely. Hence stack overflow error will occur.

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68.

What is the output of this C code?

int x = 5;
    void main()
    {
        int x = 3;
        printf("%d", x);
        {
            x = 4;
        }
        printf("%d", x);
    }

A.Run time error
B.3 3
C.3 5
D.3 4

Answer: D

Correct answer: 3 4

The inner x = 4; is an assignment, not a declaration, so it changes the x already in scope - main's local. The first printf shows 3 and the second, after the block, shows 4. The file-scope x = 5 is shadowed throughout and never seen.

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69.

What is the output of this C code?

void main()
    {
        m();
        m();
    }
    void m()
    {
        static int x = 5;
        x++;
        printf("%d", x);
    }

A.6 7
B.6 6
C.5 5
D.5 6

Answer: A

Correct answer: 6 7

A static local is initialised once, before the program starts, and keeps its value between calls - its storage lasts as long as the program even though its name is local. So the first call prints 6 and the second 7. An ordinary local would print 6 both times.

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70.

The output of the code below is

void m(int k)
    {
        printf("hi");
    }
    void m(double k)
    {
        printf("hello");
    }
    void main()
    {
        m(3);
    }

A.hi
B.hello
C.Compile time error
D.Nothing

Answer: C

Correct answer: Compile time error

C has no function overloading: two functions cannot share a name, whatever their parameters. The compiler reports a redefinition. In C++ this would compile, and m(3) would select the int version.

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