Thread: Postfix incrementer and pointer question

Because you are only passing a copy of that pointer to your function. C is pass by value. In order to actually change the value of the pointer and have those changes be visible from the caller, you need to pass a pointer to your pointer. e.g. char**.

EDIT:

Originally Posted by AndrewHunter

The confusion you are having comes down to really wrapping your head around 'pass by value'. Anything you pass to a function, including a pointer, is pass by value. Thus any changes made in that function are gone once the function returns. This includes making assignments to pointers, which is what you are doing in your second example.

A simplier example of the same concept is :

Code:

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

void foo(int* ptr){

	ptr = malloc(sizeof(int));
	*ptr = 5;
}
int main(void){

	int* mainptr = NULL;
	
	if(mainptr==NULL)
		printf("mainptr is NULL\n");
	
	foo(mainptr);
	
	if(mainptr==NULL)
		printf("mainptr is NULL\n");

	getchar();
	return(0);
}

For a similar explanation, you can read c-faq Question 4.8

Originally Posted by albundy

Perfect, thanks for the quick reply! That is what I suspected but I couldn't find the answer.

To see it more clearly... print out the value of your pointer just before you return from the function... It'll get plenty obvious then.

Afterthought: Interestingly, it's a good thing that it does work that way, otherwise your program would have crashed.