Thread: Basic pointer question

If you want to change the value of an int in a different function, you have to pass the address of the int to a function. If you want to change the value of a pointer in a different function, you have to pass the address of the pointer to the function.

I want the pointer declared in main to point to the int created on the heap when func1 returns. I thought that by passing just "val" into func1, I was passing it's address.

/sigh

I could do it this way:

Code:

int *func1(void)
{
  return malloc(sizeof(int));
}

int main()
{
  int *val = func1();
}

but I wanted to reserve the return value for an error code.

I can see what the problem is with the original code. All I'm doing is modifying the local paramter pointer var, having no effect on the one in main.

[EDIT again]

Sorry for the long posts. Maybe this will be of help to someone else. The answer lies with pointers to a pointer. This is making my head spin, but I've got it now.

Code:

#include <assert.h>
#include <malloc.h>

void func1(int **val);

int main()
{
  int *val = 0;
  func1(&val);
  assert(*val == 100);
}

void func1(int **val)
{
  int *newint = malloc(sizeof(int));
  *newint = 100;
  *val = newint;
}

Indeed. This would work (changes are commented):

Code:

#include <assert.h>
#include <stdlib.h>    // <-- malloc.h is deprecated. Use stdlib.h instead

void func1(int **val);

int main()
{
  int *val = 0;
  func1(&val);     // <---- Pass the address of val, instead of the value of val
  assert(val != 0);
}

void func1(int **val)  // <---- make val a pointer to a pointer to type int
{
  int *newint = malloc(sizeof(int));
  *val = newint;       // <----- You must now dereference val for this assignment
}

you should update func1 prototype as well
and you do not need the temporary var newint

Yeah, I forgot to update the prototype. That's fixed now. I left the temporary variable in just to avoid changing more than needed to be changed to make it work.