[C++] warning: NULL used in arithmetic

[NovaAesa]

Okay, say I have a function foo that returns a pointer to an object of type dataType. This function will either return NULL or a pointer to an object.

dataType* foo();

I want to test if the return value is NULL or not. I use code something like this:

if (foo() == NULL) {
  //do something
}

Now, the code in question does what I want it to, I'm just getting this pesky warning:

warning: NULL used in arithmetic

What's this all about? I want to fix whatever it's warning me about, but I don't see what's wrong with it.


EDIT: I can post actual code if that will help, this is just mockup of what the situation is.

[mike_g]

Looks ok to me. Are you sure its not something else thats causing the problem? Many you could post the code it came from.

On a side note i prefer using:

if(! foo())
{
  //do something
}

But then I'm lazy

[NovaAesa]

Wow, silly me. I just realised that my fuction 'foo' actually returns a boolean. I can see why I got the error now. I was comparing a boolean with a zero constant (NULL).

[Rany Albeg]

Same thing happened to me

[Paul Miller]

Isn't NULL defined as ((void*)0)? If that's the case, I don't exactly see why checking 0 == NULL triggers such a warning.

[dribeas]

if (foo() == NULL) {
  //do something
}

Standard recommendation is using 0 instead of NULL. I know that you've already solved it, but nonetheless...

[nvteighen]

Isn't NULL defined as ((void*)0)? If that's the case, I don't exactly see why checking 0 == NULL triggers such a warning.

Because they're of different types. 0 is int, (void *)0 is a pointer.