Functions Questions and Answers

Q: What is the output of this code having void return-type function? void foo() { return 1; } void main() { int x = 0; x = foo(); printf("%d", x); }

Correct answer: Compile time errorA void function must not return a value, so return 1; is an error - and assigning the result of a void function to an int is a second one. Either alone would stop the program compiling.

Q: What will be the data type returned for the following function? int func() { return (double)(char)5.0; }

Correct answer: intThe declared return type wins. Whatever casting happens inside, the value is converted to the function’s return type on the way out - so func returns an int. Here 5.0 becomes a char, then a double, and finally an int at the return.

Q: What is the problem in the following declarations? int func(int); double func(int); int func(float);

Correct answer: All of the mentionedC has no function overloading: one name may denote only one function in a translation unit. So none of the three variations is allowed - not a different signature, not a different return type, not a different number of parameters. (C++ permits overloading on the parameters, though still not on the return type alone.)

Q: The output of the code below is void main() { int k = m(); printf("%d", k); } void m() { printf("hello"); }

Correct answer: Errorm is defined after main and not declared before it, so at the call the compiler either rejects it outright or, under the old rules, assumes int m() - and the later definition as void m() then conflicts with that assumption. Either way the program does not compile. Assigning the result of a void function to an int would be an error in its own right.

Q: The output of the code below is int *m() { int *p = 5; return p; } void main() { int *k = m(); printf("%d", k); }

Correct answer: 5int *p = 5; puts the value 5 into a pointer - meaningless as an address, but the bit pattern is simply carried back and printed with %d as 5. The code compiles only with warnings, and would crash the moment anything dereferenced p.

Q: The output of the code below is int *m(); void main() { int *k = m(); printf("hello "); printf("%d", k[0]); } int *m() { int a[2] = {5, 8}; return a; }

Correct answer: hello followed by garbage valuea is a local array, and it ceases to exist when m returns - so the pointer returned is dangling. Reading k[0] reads memory that has been released and may already have been reused, giving whatever happens to be there. To return an array safely it must be static, or allocated with malloc, or supplied by the caller.

Q: The output of the code below is int *m(); void main() { int k = m(); printf("%d", k); } int *m() { int a[2] = {5, 8}; return a; }

Correct answer: VariesTwo faults compound each other. The function returns a pointer to a local array that no longer exists, and the caller assigns that pointer to an int. What is printed is an address reinterpreted as an integer - different on every run and every machine.

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.

Q: What is the default return type if it is not specified in function definition?

Correct answer: intUnder the older rules an omitted return type defaults to int. C99 removed the implicit int, so a modern compiler warns or refuses - but the default, where it is honoured, is int.

Q: What is the output of this C code? int foo(); int main() { int i = foo(); } foo() { printf("2 "); return 2; }

Correct answer: 2The prototype says int foo(), and the definition, which gives no return type, defaults to int - so the two agree. The program compiles and prints 2.

61.

What is the output of this code having void return-type function?

void foo()
    {
        return 1;
    }

    void main()
    {
        int x = 0;
        x = foo();
        printf("%d", x);
     }

A.1
B.Runtime error
C.Compile time error

Answer: C

Correct answer: Compile time error

A void function must not return a value, so return 1; is an error - and assigning the result of a void function to an int is a second one. Either alone would stop the program compiling.

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62.

What will be the data type returned for the following function?

int func()
    {
        return (double)(char)5.0;
    }

A.char
B.int
C.double
D.multiple type-casting in return is illegal

Answer: B

Correct answer: int

The declared return type wins. Whatever casting happens inside, the value is converted to the function’s return type on the way out - so func returns an int. Here 5.0 becomes a char, then a double, and finally an int at the return.

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63.

What is the problem in the following declarations?
     int func(int);
     double func(int);
     int func(float);

A.A function with same name cannot have different signatures
B.A function with same name cannot have different return types
C.A function with same name cannot have different number of parameters
D.All of the mentioned

Answer: D

Correct answer: All of the mentioned

C has no function overloading: one name may denote only one function in a translation unit. So none of the three variations is allowed - not a different signature, not a different return type, not a different number of parameters. (C++ permits overloading on the parameters, though still not on the return type alone.)

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64.

The output of the code below is

void main()
    {
        int k = m();
        printf("%d", k);
    }

    void m()
    {

        printf("hello");
    }

A.hello 5
B.Error
C.Nothing
D.Junk value

Answer: A

Correct answer: Error

m is defined after main and not declared before it, so at the call the compiler either rejects it outright or, under the old rules, assumes int m() - and the later definition as void m() then conflicts with that assumption. Either way the program does not compile. Assigning the result of a void function to an int would be an error in its own right.

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65.

The output of the code below is

int *m()
    {
        int *p = 5;
        return p;
    }
    void main()
    {
        int *k = m();
        printf("%d", k);
    }

A.5
B.7
C.Junk value

Answer: A

Correct answer: 5

int *p = 5; puts the value 5 into a pointer - meaningless as an address, but the bit pattern is simply carried back and printed with %d as 5. The code compiles only with warnings, and would crash the moment anything dereferenced p.

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66.

The output of the code below is

int *m();
    void main()
    {
        int *k = m();
        printf("hello ");
        printf("%d", k[0]);
    }
    int *m()
    {
        int a[2] = {5, 8};
        return a;
    }

A.hello 5 8
B.hello 5
C.hello followed by garbage value
D.Compilation error

Answer: C

Correct answer: hello followed by garbage value

a is a local array, and it ceases to exist when m returns - so the pointer returned is dangling. Reading k[0] reads memory that has been released and may already have been reused, giving whatever happens to be there. To return an array safely it must be static, or allocated with malloc, or supplied by the caller.

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67.

The output of the code below is

int *m();
    void main()
    {
        int k = m();
        printf("%d", k);
    }
    int *m()
    {
        int a[2] = {5, 8};
        return a;
    }

A.5
B.8
C.Nothing
D.Varies

Answer: D

Correct answer: Varies

Two faults compound each other. The function returns a pointer to a local array that no longer exists, and the caller assigns that pointer to an int. What is printed is an address reinterpreted as an integer - different on every run and every machine.

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68.

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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69.

What is the default return type if it is not specified in function definition?

A.void
B.int
C.double
D.short int

Answer: B

Correct answer: int

Under the older rules an omitted return type defaults to int. C99 removed the implicit int, so a modern compiler warns or refuses - but the default, where it is honoured, is int.

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70.

What is the output of this C code?

int foo();
    int main()
    {
        int i = foo();
    }
    foo()
    {
        printf("2 ");
        return 2;
    }

A.2
B.Compile time error
C.Depends on the compiler
D.Depends on the standard

Answer: A

Correct answer: 2

The prototype says int foo(), and the definition, which gives no return type, defaults to int - so the two agree. The program compiles and prints 2.

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