Passing Pointer to a function

I am not getting the obvious output (67, 67) for this program. Why I don't get the value of the local variable of a fucntion to which i am passing a pointer. I can pass the value of a variable to a function through pointer , but not receive the value from a function through a pointer.


#include "stdio.h"

void func(int *p)
{
int i= 67;
p = &i;
printf("\nIn func() %d\n",*p);
}

void main(void)
{
int *ptr;
func(ptr);
printf("\nIn main() %d\n",*ptr);
}



Thanks,
Nishant Ghai

Quote:

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Originally posted by Nishant Ghai
I am not getting the obvious output (67, 67) for this program. Why I don't get the value of the local variable of a fucntion to which i am passing a pointer. I can pass the value of a variable to a function through pointer , but not receive the value from a function through a pointer.


#include "stdio.h"

void func(int *p)
{
int i= 67;
p = &i;
printf("\nIn func() %d\n",*p);
}

void main(void)
{
int *ptr;
func(ptr);
printf("\nIn main() %d\n",*ptr);
}



Thanks,
Nishant Ghai

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First, don't void your mains, ... int them
main should return an int.
Second, scope of the variable i, is limited to the function fun, outside of that it has no meaning neither has its address.
Third,ptr is a pointer, and is passed by value, its value in main is not changed just by passing it in a function, you can possibly change the value of what a pointer points to this way but not the pointer itself.

you could probably fix all your problems by making *ptr outside of the main function as a global variable, but that's just the lazy way :D . With any program bigger than that one it would of course be better to restructure your functions to fix the problem

Well when func ends, i goes out of scope, so ptr isn't valid anymore.

If you really want to do this...

Code:

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#include "stdio.h"

void func(int **p)
{
  static int i= 67;
  *p = &i;
  printf("\nIn func() %d\n",**p);
}

int main(void)
{
  int *ptr;
  func(&ptr);
  printf("\nIn main() %d\n",*ptr);
  return 0;
}

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