How is sizeof(*NULL) equal to 1 on 32-bit C compiler?

Question

I recently learned that sizeof is not a function and actually evaluates at compile time.

But I still haven't been able to figure out how sizeof(*NULL) is equal to 1.

EDIT: I'm talking about gcc here specifically.

Answer 1

It depends on the compiler and standard library. With GCC and glibc, NULL is defined as (void*)0 and GCC has an extension that allows pointer arithmetic on void pointers using sizeof(void) == 1. Without this extension, the compiler would produce an error.

A standard library could also define NULL as 0. In this case sizeof(*NULL) would also be illegal.

Answer 2

Standard makes no guarantee as to the value returned by sizeof(*NULL), which is enough to not use this construct in practice.

Your code uses implementation-defined behavior.

7.173 The macros are

NULL

which expands to an implementation-defined null pointer constant

This means that the standard does not guarantee that your code is going to compile, let alone producing a certain value.

On systems defining macro NULL as ((void*)0) your code will trigger a warning that should be treated as an error:

prog.c:4:28: error: invalid application of ‘sizeof’ to a void type [-Werror=pointer-arith]

Demo.