Comparing size of NULL macro with size of char

Question

I'm testing to see that the NULL constant does indeed occupy the size of a pointer:

ASSERT(sizeof NULL == sizeof(char*));

However, I accidentally wrote the following instead:

ASSERT(sizeof NULL == sizeof char);

That should have compiled, but instead it gave me the following error:

error: expected expression before ‘char’

The same happened after I enclosed NULL in brackets

ASSERT(sizeof(NULL) == sizeof char);

Isn't the NULL constant usually suppose to be defined by a macro which associates it to a pointer which is equal to 0? The statement was obviously false but as far as I see there was no syntactic error. If that is true, why was I receiving a compilation error?

Type names must be enclosed in parentheses when they are the operand of `sizeof`. — Filipe Gonçalves, Mar 02 '14 at 12:32
`NULL` may be defined as `0` or `(void*)0` or otherwise. There is no guarantee that it will have the same size as a pointer. — interjay, Mar 02 '14 at 12:36
sizeof() requires parenthesis per http://stackoverflow.com/questions/5894892/why-do-i-need-to-use-parentheses-after-sizeof — abligh, Mar 02 '14 at 12:40
@PranitKothari C allows `NULL`, a _null pointer constant_, to be an `int` or a `void*`. "An integer constant expression with the value 0, or such an expression cast to type void *, is called a null pointer constant.". As I read that, `NULL` could also be `unsigned` or `long long`, etc. — chux - Reinstate Monica, Aug 17 '18 at 15:59

score 1 · Accepted Answer · answered Mar 02 '14 at 12:44

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"When the operand is a type name, it must be enclosed in parentheses": C sizeof operator

In C, NULL is usually defined as

#define NULL ((void*)0)

answered Mar 02 '14 at 12:44

Spock77

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I think you mean `#define NULL ((void*)0)` – Filipe Gonçalves Mar 02 '14 at 12:45
@FilipeGonçalves - yep, sure. – Spock77 Mar 02 '14 at 12:46

Comparing size of NULL macro with size of char

1 Answers1