Passing NULL as argument to a C macro - c

Question

I am trying to define a macro -
#define macro1(arg1) \
do{ \
int _state = 0; \
if (arg1 && arg1->member_) \
_state = arg1->member_->state_; \
printf("%d", _state); \
} while(0)
A *a = new A():
macro1(a); // Works
macro1(NULL); // Error
The specific error I see is -
"error: base operand of ‘->’ is not a pointer"
Aren't we allowed to pass NULL as an argument to macros?

Answer 1

Macro expansion is just text replacement, so when you passed NULL, it will expand to NULL->member, clearly it is an error. One way is to use a temporary variable for that:
#define macro1(arg1) \
do{ \
A* p = (arg1);
int _state = 0; \
if (p && p->member_) \
_state = p->member_->state_; \
printf("%d", _state); \
} while(0)
A *a = new A():
macro1(a);
macro1(NULL);
This way both cases will work.

Answer 2

You have to understand what's a macro in order to understand your mistake. Except for the compiler, there's an animal called pre-compiler. It replaces all the macros' references by the actual code defined for this macro. So this code:
#define macro1(arg1) \
do{ \
int _state = 0; \
if (arg1 && arg1->member_) \
_state = arg1->member_->state_; \
printf("%d", _state); \
} while(0)
A *a = new A():
macro1(a); // Works
macro1(NULL); // Error
will be replaced with:
A *a = new A():
do{
int _state = 0;
if (a && a->member_)
_state = a->member_->state_;
printf("%d", _state);
} while(0)
do{
int _state = 0;
if (NULL && NULL->member_)
_state = NULL->member_->state_;
printf("%d", _state);
} while(0)
THIS code will be compiled. And now you can see for yourself what's the root cause of the compilation error.

Answer 3

Macros are just a text replacement.
for example, if you have
#define mac(x) x/x
that would work for must numbers but not for 0, because it will be replaced with 0/0 which is not defined.
in your case if you pass NULL it will be replaced with:
do{ \
int _state = 0; \
if (NULL && NULL->member_) \
_state = NULL ->member_->state_; \
printf("%d", _state); \
} while(0)
so what is the meaning of NULL->member_ in this case. No sense, hence it fails.
consider using a regular function, or two macros one for regular pointers and one for NULL pointers, and make your code as this:
if (ptr)
macro1(ptr);
else
macro2;

Answer 4

Small adjust let the pre-compiler know the type size + a forward declaration and it'll work:
#define macro1(arg1) \
do{ \
int _state = 0; \
if ((arg1) && ((A*)arg1)->member_) \
_state = ((A*)arg1)->member_->state_; \
printf("%d", _state); \
} while(0)
Complete code:
#include <stdio.h>
class A;
#define macro1(arg1) \
do{ \
int _state = 0; \
if ((arg1) && ((A*)arg1)->member_) \
_state = ((A*)arg1)->member_->state_; \
printf("%d", _state); \
} while(0)
struct member{
int state_;
};
class A {
public:
member* member_;
};
int main(int n, char** arg) {
A* a = new A();
a->member_ = new member();
a->member_->state_ = 1;
macro1(a);
macro1(NULL);
return 0;
}

Answer 5

The more fundamental problem is that NULL is not a pointer, it is a macro for 0.
Hence when you pass in NULL, it is equivalent to passing in 0, which of course is an error.
As other answers have mentioned, giving the argument an explicit cast will fix it
(A*)arg1
As a side note, you should be using nullptr in C++ for a null pointer.
EDIT: As #AjayBrahmakshatriya pointed out, NULL can be defined as (void*)0 (in C only) which will be a pointer, but the argument still holds true, NULL is not a pointer of your type.
EDIT2: Apparently in C++11 and later, NULL can be defined as nullptr