Don’t use NULL

I always thought that using NULL whenever I wanted to assign a null pointer value to a pointer was a good thing, but today I learned the contrary.

The standard, 6.3.2.3, says, seemingly innocent

An integer constant expression with the value 0, or such an expression cast to type void*, is called a null pointer constant.

An integer constant expression is something that evaluates at compile time and is of an integer type. Integer types are all the standard integer types that we all know, including plain char and _Bool, and enumeration types. In particular such expressions may contain casts to such types.

So far this all sounds relatively innocent if you think of the primary use of null pointer constants namely to be assigned to pointer variables, but see below.

Then later (7.17) is defined the macro

NULL which expands to an implementation-defined null pointer constant

This means that NULL may result in at least 12 different standard integer types and many more expressions that result in a null value in that type:

(_Bool)0
(char)0
(unsigned char)0
(signed char)0
(unsigned short)0
(signed short)0
(unsigned)0 == 0U
(signed)0 == 0 == '\0' == enumeration_constant_of_value_0
(unsigned long)0 == 0UL
(signed long)0 == 0L
(unsigned long long)0 == 0ULL
(signed long)0 == 0LL

and in

(void*)0

and for any enumeration type enum etag or any typedef tdef of integer type in

(enum etag)0, (tdef)0.

What is important here that most of these expressions are only interpreted as null pointer constants when they appear in what some people call a pointer context that is in a context where the compiler expects a pointer value. But strictly speaking C, doesn’t know such a thing like a pointer or integer context.

This becomes harmful when you pass arguments to va_arg functions. Here most of the arguments are just placed on the stack as they are, only small integer types are promoted to int. Now consider a harmless function such as printf that at some point expects a void* pointer:

printf("%d (%p) %d", 1, NULL, 2);

Looks fine? But it isn’t at all. As we have seen depending on the mood of your compiler provider NULL here may be many things: something what most people expect, namely (void*)0, but also 0L or 0LL. The call to printf might simply expect the three arguments on wrong places on the stack and crash your program.

Now people seriously propose the following work around for that

printf("%d (%p) %d", 1, (void*)NULL, 2);

But wasn’t the only justification of the macro NULL just that it should remind us that there is a pointer type expected at a certain place? But this fact is already clear from the cast to void*. Why not just use

printf("%d (%p) %d", 1, (void*)0, 2);

that has the same message?

In summary,

• NULL obfuscates an expression that can be either of integer type or of pointer type.
• Whenever you want to initialize or assign to a pointer to a null value, use plain 0, it does serve well.
• Whenever you use a null value in a function call for a parameter that has no prototype use (T*)0.

Ah, and yes, an enumeration casted 0‘s should be interpreted as null pointer constants:

enum awk { a };
void* q = (enum awk)0;

clang accepts this correctly. gcc and opencc don’t.