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 4 of 10.

Kalyan said:   1 decade ago
As p is the void pointer so we have to convert to an integer pointer so we made the conversion. And extern varible default value is 0.

Nandkishor said:   10 years ago
int i; is declare as external. Hence for external storage class default value is 0. That is the reason why it prints 0.

Rachit said:   1 decade ago
Please tell me the significance of (int**) because if we write (int) or (int****...) then also the answer is same.

Preethi said:   1 decade ago
What does this line mean actually?.

q = (int**) &p;.

Can anyone explain me please?.

Ashish said:   1 decade ago
That's because I is declared as a global and and not been initialized. Default value of global variable is 0.

Ch.suresh said:   1 decade ago
I think void *p represent a null pointer so , null pointer means it pointing to null=0.
eg: *p->0(null)

Dpol.. said:   1 decade ago
Is I extern variable as extern is not mentioned. It should be automatic according to me? please elaborate.

SREE said:   1 decade ago
I think void *p represent a null pointer so, null pointer means it pointing to null=0.
eg: *p->0(null)

Haripriya Joshi said:   1 decade ago
As we all are seeing that i is declared as global and the default initial value of global variable is 0.

Divya said:   8 years ago
Local variable will initialize garbage values, as well as global variables will initialize as 0.


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