C Programming - Pointers - Discussion

Discussion Forum : Pointers - Find Output of Program (Q.No. 6)
6.
What will be the output of the program ?
#include<stdio.h>

void fun(void *p);
int i;

int main()
{
    void *vptr;
    vptr = &i;
    fun(vptr);
    return 0;
}
void fun(void *p)
{
    int **q;
    q = (int**)&p;
    printf("%d\n", **q);
}
Error: cannot convert from void** to int**
Garbage value
0
No output
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
95 comments Page 6 of 10.

Paddu said:   1 decade ago
Answer is 0 because i is external variable.

Default value of external variable is 0.

Annapurna said:   1 decade ago
Can we declare void as a datatype for parameters and how the compiler treats it?

Sumit kumar nager said:   1 decade ago
i is a global variable so its value is 0 by default.

vptr = &i; \\we can assign any kind of address to void type variable.

q = (int**)&p; \\ this is type casting void pointer is converted into int

So at the end we have
i=0;
q=0;

Sunil said:   1 decade ago
i is declare as global variable & default value of global variable is 0.

Ankitradhe said:   1 decade ago
Global variable default value is 0.

Sujay said:   1 decade ago
Can any one explain me about (int**) &p;?

Jerin said:   1 decade ago
in (int**)p p is typecasted to integer pointer.

Seema said:   1 decade ago
As i is declared global and default value of global variables is 0.

Sachin said:   1 decade ago
@Seema.

i is declared global its default value is 0 as well as it is stored in bss(block starting with symbol);

One more thing is that if it was initialized then it will be stored on the data section of memory.

Swapnil said:   1 decade ago
i is neither static nor extern. It's a variable visible for the compilation unit it's in, and additionally will be visible from all compilation units that declare x to be an extern variable.

Why am I saying it's neither static nor extern?

If it was extern, then, there must be a different compilation unit with x declaration on it. Clearly this is your only compilation unit.

If it was static then, no extern reference would be allowed to x variable defined in this compilation unit. We know that we could easily declare an extern variable to this x declared here.

Why is 0 assigned to x? Because, in C, all global variables initialize to 0. It says so in 6.7.8 (10) of the C99 standard.


Post your comments here:

Your comments will be displayed after verification.