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 5 of 10.

Abs said:   1 decade ago
Because i is declared as extern and the default value of an extern variable is 0 ... i think :)

Shashank said:   1 decade ago
The value refer to the pointer p& as the q refer to the extern variable, so the value is 0.

Sachendra said:   1 decade ago
@Anil ,i we don't declare i as a global then it will print garbage value that is stored in i.

Sivakumarn said:   1 decade ago
I don't know how integer pointer convert to null pointer any body please can you explain.

Indira said:   1 decade ago
Can anyone tell me?

q = (int**)&p = q = (int)&p.

What is the use of (int**).

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

Default value of external variable is 0.

Kalyan kedarnath said:   5 years ago
q=(int**)&p // Typecasting.

i initial value is Zero 0 because Global variable.
(1)

Piyush said:   1 decade ago
Hi guys,

Its answer is 0, because I is declare as global. So its by default zero.

Maheswari said:   8 years ago
If we initialized i=5 it will print 5. Whatever I value it will print that value.

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


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