Thread: pass by reference question

I know C doesn't have a "true" pass by reference, but I also know it can be simulated by using pointers.
While it works I want to know if the following would be the best way to pass a variable recieved by reference to another routine by refence. (subroutine1 passes var x by ref to subroutine2 which then passes to subroutine3 by ref).

Code:

void sub1(int x)
{
  sub2(&x);
}

void sub2(int* x);
{
  sub3(&*x);
}

void sub3(int* x);
{
}

I know the code I posted does jack squat but I'm more interested in making sure that the way I did the passing is proper. (real code does some addition and printf to show that it does work)

Not sure that I'm following ya Sebastiani. Here is the actual code I used to test (minus the other combos).

Code:

#include <stdio.h>

void print(int, char*, int, int);

void test7(int*);
void test8(int*);

int main (void)
{
	int x=4;
	printf("\nR -> R\t%d\n",x);
	test7(&x);
	printf("At end: %d\n", x);
}

void print(int a, char* s, int b, int c)
{
	printf("test%d %s test%d : %d\n", a, s, b, c);
}

void test7(int* x)
{
	print(7, "Before Inc", 7, (int) *x);
	++(*x);
	print(7, "Before", 8, (int) *x);
	test8(&*x);
	print(7, "After", 8, (int) *x);
}

void test8(int* x)
{
	print(8, "Before Inc", 8, (int) *x);
	++(*x);
	print(8, "After Inc", 8, (int) *x);
}

Which gave me the output of:

Code:

R -> R  4
test7 Before Inc test7 : 4
test7 Before test8 : 5
test8 Before Inc test8 : 5
test8 After Inc test8 : 6
test7 After test8 : 6
At end: 6

This to me seems to indicate that it works.
I've tried test8(&x) and test(*x) and I always get a compile error.

The following works for me. I changed the variable name in sub2 and sub3 from x to xp. This reflects the fact that it is actually an integer pointer that points to where x is stored. I find this type of naming always helps me.

Code:

#include <stdio.h>
#include <stdlib.h>
#include <stdio.h>

void sub3(int* xp)
{
  printf("4) in sub3 *xp = %d\n",*xp);  
  *xp=6;
  printf("5) in sub3 *xp = %d\n",*xp);  
}

void sub2(int* xp)
{

  printf("3) in sub2 *xp = %d\n",*xp);
  sub3(xp);
  printf("6) in sub2 *xp = %d\n",*xp);
}

void sub1(int x)
{
  printf("2) in sub1 x = %d\n",x);
  sub2(&x);
  printf("7) in sub1 x = %d\n",x);
}

int main(void)
{
  int x = 5;
  
  printf("1) in main x = %d\n",x);
  sub1(x);
  printf("8) in main x = %d\n",x);
    
  system("PAUSE");	
  return 0;
}

Thanks Prelude! I understand it clearly now.