Pointer arithmetic when void has unknown size [closed]

Answer 1

In C language sizeof cannot be applied to incomplete types. void is an incomplete type, which is why you can't use it as operand of sizeof.

Pointer arithmetic of void * pointers is also not supported in C. In order to be used with pointer arithmetic the pointer has to point to object type, which void is not.

GCC allows both as a weird language extension.

Answer 2

void * is well recognised and used as a typeless pointer to raw memory.

Correct. Not having a type means not having a size either.

Answer 3

It would appear the correct answer is to use char * for pointer arithmetic, because sizeof(char) is always defined to be 1, and to be of the finest addressable granularity on any platform.

So in short, there is no way around the limitation, char * is in fact the proper way to do it.

Matthias Wandel had the right answer but with a different justification.

Answer 4

void * is meant to imply "pointer to unknown"

Adding 6 to a void pointer is therefore undefined, because you're asking the compiler to advance the pointer by "the size of 6 unknown objects".

You should not be doing it.

Answer 5

I can see the error on the first line at least - you have sizeof(void) and it should be sizeof(void*), no?

Answer 6

Incrementing a pointer is technically supposed to increment it by whatever size of thing that its pointing at. Void points at nothing, so the size is unknown.

Its a nicety for some compilers to let you increment void pointers.

Casting is annoying, because you have to either cast it to another pointer, and then back, or do something awkward like (*(char *)&voidptr) += 6

Personally, I'd just declare it as a char * for purposes of arithmetic