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);
}
Discussion:
95 comments Page 7 of 10.
Shambhu said:
1 decade ago
In this program i is declared as a global variable. And we know that global variable initially initialized to be 0.
So int i=0;
Now void pointer(it is a pointer which points any data type)pointing to the base address of int i.
So it point the base address of i(means 0).
1st step:
vptr contain the address of i, so it contain 0.
and passes it through function.
2nd step:
In fun....
We assign the value contain in *p in q.
So dereference it we get the value present in i.
i.e----> 0.
So int i=0;
Now void pointer(it is a pointer which points any data type)pointing to the base address of int i.
So it point the base address of i(means 0).
1st step:
vptr contain the address of i, so it contain 0.
and passes it through function.
2nd step:
In fun....
We assign the value contain in *p in q.
So dereference it we get the value present in i.
i.e----> 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.
Anu said:
1 decade ago
void *vptr;
Is it correct?
Is it correct?
Raj Naik said:
1 decade ago
@Anu : yeah..void *vptr is valid.
A void pointer is pointer which has no specified data type and the void pointer can be pointed to any type. If needed, the type can be casted as : int **q =(int **)&vptr.
A void pointer is pointer which has no specified data type and the void pointer can be pointed to any type. If needed, the type can be casted as : int **q =(int **)&vptr.
Akshay K said:
1 decade ago
In given code, vptr is a void pointer and 'i' is integer variable.
But vptr assigned the address of integer variable without typecasting.
see code:
---------------
int main()
{
void *vptr;
vptr = &i;
--------------
So, it takes value as 0.
But vptr assigned the address of integer variable without typecasting.
see code:
---------------
int main()
{
void *vptr;
vptr = &i;
--------------
So, it takes value as 0.
Indira said:
1 decade ago
Can anyone tell me?
q = (int**)&p = q = (int)&p.
What is the use of (int**).
q = (int**)&p = q = (int)&p.
What is the use of (int**).
Shruthi said:
1 decade ago
Whether *(int*) is same as (int**)?
Vicky said:
1 decade ago
Since it has the type void it is called as null pointer. Which is general purpose pointer. Instead of declaring pointer as int, float, char as 3 times.
We can use null pointer. Which stores address of any datatype for eg: int, float.
Because it does not have any type, compiler does not know what datatype pointer is holding. For that we have to do typecast.
int main()
{
int inum = 8;
float fnum = 67.7;
void *ptr;
ptr = &inum;
printf("\nThe value of inum = %d",*((int*)ptr));//tells compiler that ptr is holding integer value and returns 8
ptr = &fnum;
printf("\nThe value of fnum = %f",*((float*)ptr));//tells compiler that ptr is holding float value and returns 67.7
return(0);
}
We can use null pointer. Which stores address of any datatype for eg: int, float.
Because it does not have any type, compiler does not know what datatype pointer is holding. For that we have to do typecast.
int main()
{
int inum = 8;
float fnum = 67.7;
void *ptr;
ptr = &inum;
printf("\nThe value of inum = %d",*((int*)ptr));//tells compiler that ptr is holding integer value and returns 8
ptr = &fnum;
printf("\nThe value of fnum = %f",*((float*)ptr));//tells compiler that ptr is holding float value and returns 67.7
return(0);
}
Mohamed mohsen said:
1 decade ago
Because i is global variable by default compiler initialize it to zero.
Prasun said:
1 decade ago
By default the global variable has default value 0. So answer is 0.
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