Thread: passing value or reference?

hi all,
i have a doubt for this simple program. i'm passing just a copy of p to changeNum function. why is it changing the original? i know its pretty silly but i can't understand this. please tell me. One more question is if ihave int **ptr, how to pass ptr to a function so it can change the value. thanks in advance.

Code:

#include <stdio.h>

void changeNum(int *);

int main(void)
{
	int i = 0;
	int *p;

	p = &i;
	printf("*p is %d\n", *p);
	changeNum(p);
	printf("*p is %d\n", *p);
	return 0;
}

void changeNum(int *p)
{
	*p = 10;
}

i found this on website. please, what does this mean?
http://www.eskimo.com/~scs/C-faq/q4.8.html

Question 4.8
I have a function which accepts, and is supposed to initialize, a pointer:

Code:

void f(ip)
int *ip;
{
	static int dummy = 5;
	ip = &dummy;
}

But when I call it like this:

Code:

int *ip;
f(ip);

the pointer in the caller remains unchanged.


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Are you sure the function initialized what you thought it did? Remember that arguments in C are passed by value. The called function altered only the passed copy of the pointer. You'll either want to pass the address of the pointer (the function will end up accepting a pointer-to-a-pointer), or have the function return the pointer.

thank you both. i got it!