C++ invalid use of void, why?

Question

char mem[8];
uint64_t *memory{(uint64_t*)(void*)&mem[0]};
std::cout << "diff: " << (void*)memory - (void*)(&mem[0]) << std::endl;

Trivial example, error message with gcc is:

error: invalid use of ‘void’
std::cout << "diff: " << (void*)memory - (void*)(&mem[0]) << std::endl;

What are you trying to do? reinterpret the `char[]` as a single `uint64_t`? — Cory Kramer, Aug 10 '17 at 16:41
@CoryKramer Yeah, pretty much, trying to convert a `char[]` to a `uint64_t` array. I thought it would be easy, but this has got me stumped... — FreelanceConsultant, Aug 10 '17 at 16:43
VS2013 shows error - `void*:unknown size`. Expression must be a pointer to complete object type. — Nipun, Aug 10 '17 at 16:44

ikleschenkov · Answer 1 · 2017-08-10T16:49:17.467

1

invalid use of void, why?

It's because size of void* (item that stored at this pointer) is unknown .

If you want to calculate distance between pointers in bytes, cast them to pointers of something 1-byte-sized (char*, unsigned char*, uint8_t*, etc.):

std::cout << "diff: " << (uint8_t*)memory - (uint8_t*)(&mem[0]) << std::endl;

edited Aug 10 '17 at 16:49

answered Aug 10 '17 at 16:43

ikleschenkov

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What happens if I convert them to `(uint64_t*)` by the way? My assumption would be the difference would still be the same, measured in byte? – FreelanceConsultant Aug 10 '17 at 16:44
@user3728501 it will measure the distance in the `uint64_t` (8bytes) rounded to bottom. For example, distance `2 bytes` will be the `0 uint64_t` – ikleschenkov Aug 10 '17 at 16:46
"distance 2 bytes will be the 0 uint64_t" - this doesn't make any sense? – FreelanceConsultant Aug 10 '17 at 17:38
@user3728501 well, it's because it's wrong to count in `uint64_t` when you store `uint8_t` at pointer. Use 1-byte pointers to calc the size (distance) in bytes and pointer to the actual stored items (plain types, objects, structures) to calculate items count. – ikleschenkov Aug 10 '17 at 19:40

score 0 · Answer 2 · answered Aug 10 '17 at 16:43

You can't perform pointer arithmetic with void pointers - see here. This is due to void not having a defined "size", whereas eg. with int *x, you know that x+1 is going to be sizeof(int) bytes further along that the address of x.

You can also test it here - removing either of the expressions on the sides of the - will let it compile.

C++ invalid use of void, why?

2 Answers2