How to understand accessing NULL pointer when using in sizeof?

Question

Check this C program:

#include <stdio.h>

int main(void) {
        // your code goes here
        char **p = NULL;

        printf("%d,%d\n", sizeof(*p), sizeof(**p));
        return 0;
}

Executing the above code, the result is:

8,1

Although p is NULL, it doesn't cause program crash in sizeof(*p) and sizeof(**p). How to understand this behavior? Is it assured in c spec?

If you think about what is happening sizeof(*p) is just sizeof (char *) and sizeof(**p) is just sizeof(char) — David C. Rankin
– David C. Rankin, Commented Apr 27, 2016 at 6:35

Lundin · Accepted Answer · 2016-04-27 06:41:08Z

5

The sizeof operator is evaluated at compile time. Its operand is not evaluated for side effects, so your program is safe. This is guaranteed by the standard 6.5.3.4/2 (emphasis mine):

If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant.

(Note that there is a special case of variable length arrays, in which case the evaluation takes place in run time, so code de-referencing an invalid pointer to a VLA inside sizeof would not be safe.)

As a side note, the correct format specifier for printf when printing the result of sizeof is %zu (the result of sizeof is type size_t).

edited Apr 27, 2016 at 6:41

answered Apr 27, 2016 at 6:35

Lundin

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WiSaGaN · Accepted Answer · 2016-04-27 06:35:09Z

4

Because sizeof(exp) is a compile time operator, and it does not evaluate expression exp at run-time.

As a result, there is no dereference of NULL pointer at run-time. You just have equivalent machine code of a constant in your printf statement in your final binary.

edited Apr 27, 2016 at 6:35

answered Apr 27, 2016 at 6:33

WiSaGaN

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John3136 · Accepted Answer · 2016-04-27 06:35:24Z

3

You know p is a char** so sizeof(*p) == sizeof(char*) - no dereference is actually required.

Same applies to sizeof(**p) - doesn;t have to do any dereference as it can figure out the size at compile time.

answered Apr 27, 2016 at 6:35

John3136

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user6259831 · Accepted Answer · 2016-04-27 06:32:39Z

2

The operand of sizeof is unevaluated. In other words, p is never deferenced, therefore it isn't undefined behavior. On a 64-bit system, a pointer will be 8-bytes wide, and a char is always 1 byte, explaining your output.

answered Apr 27, 2016 at 6:32

user6259831

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