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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.
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NULL in C is defined as (void*)0. Since it's a pointer, it takes 4 bytes to store it. And, "" is 1 byte because that "empty" string has EOL character ('\0').
32 bit architectures are word aligned and byte referenced. It is irrelevant how the architecture stores a char, but to compiler, you must reference chars 1 byte at a time, even if they use up less than 1 byte. This is why sizeof(char) is 1.
Since NULL is defined as ((void*)0), we can think of NULL as a special pointer and its size would be equal to any pointer. If the pointer size on a platform is 4 bytes, the output of the above program would be 4.
In C, NULL is a macro that expands either to 0 or (void*)0 (or something that has a similar effect). In the first case, you can not differentiate between NULL and 0 , because they are literally the same.
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.
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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
NULLwhich 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.
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