Explain This Pointer

[pobri19]

Hi, I'm just after a bit of clarification here. Been fiddling with pointers, trying to get familiar with them and understand exactly what they do and how they operate. Here is the code:

#include <stdio.h>

int main()
{
	int *pCake;
	int dog;
	dog = 25;
	pCake = &dog;
	printf("Should return 25 / pCake dereferenced(*pCake): %d\n", *pCake);
	printf("Prints dog's address %d\n", &dog);
	printf("Should return dog's address (pCake): %d\n", pCake);
	printf("Print pCake's address: %d\n", &pCake);
	return 0;
}

This code returns this when run:

Should return 25 / pCake dereferenced(*pCake): 25
Prints dog's address -1074293652
Should return dog's address (pCake): -1074293652
Print pCake's address: -1074293648

( ^^^ This is the address at which the pointer(pCake is located?).

Now as far as I can see everything is written correctly, and I'm pretty confident it's right, I'm just wondering if I'm on the right track :P. Is there anything wrong with what I've done, are the address correct? Also, why are the addresses followed by a hyphen "-"?
Thanks.

[MacGyver]

You're right so far with your theory. You should be printing addresses with %p, however, to be entirely sure you're getting the right values. The numbers are showing with '-' because you're printing them as signed integers, and they are coming out negative because they are too high in value.

[laserlight]

Also, the %p format specifier is meant to be used with pointers to void, so you should actually cast your addresses to (void*) when printing them.

[pobri19]

Also, the %p format specifier is meant to be used with pointers to void, so you should actually cast your addresses to (void*) when printing them.

Sorry I'm not quiet sure what you mean, but I understand why I need %p instead of %d ;p.

[laserlight]

As in you should write:

printf("Prints dog's address %p\n", (void*)&dog);
printf("Should return dog's address (pCake): %p\n", (void*)pCake);

[matsp]

Sample of changes needed.

	printf("Print pCake's address: %p\n", (void *)&pCake);

As stated, the address of a pointer may well be above the magical "negative above this line" mark. In your case, I suspect you are using Linux or Unix, which makes the address end up just under 3GB, 0xC0000000 (e.g. 0xBFFF0000).

Ah, noticed that Laserlight just posted the same thing.

--
Mats

[pobri19]

Yeah Linux, gcc to compile.

Laser, what advantage does having that do? And how will it effect the way my program works if I don't use it?

[matsp]

Yeah Linux, gcc to compile.

Laser, what advantage does having that do? And how will it effect the way my program works if I don't use it?

The benefit is that you do NOT rely on undefined behaviour. The %p format is only guaranteed to work on void pointers. It MAY work on other pointers - but that is entirely up to the C runtime, compiler and processor architecture.

--
Mats

[laserlight]

Laser, what advantage does having that do? And how will it effect the way my program works if I don't use it?

If you do not do that, the behaviour is undefined, even though typically it will do what you expected.