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Re: Why is size_type in std::array an alias for size_t?

From: Lyberta <lyberta_at_[hidden]>
Date: Sat, 09 Nov 2019 23:07:00 +0000
Bo Persson via Std-Discussion:
> On 2019-11-09 at 12:41, Wilhelm Meier via Std-Discussion wrote:
> But if CHAR_BIT is 10, you probably don't have any char8_t or char16_t
> types. So in a way it *does* matter.

char8_t and char16_t will still exist because char8_t has the same
representation as unsigned char and char16_t has the same representation
as std::uint_least16_t. All those types are mandatory.

std::[u]int[8,16,32,64]_t won't exist though as they are optional and
fixed size.

Received on 2019-11-09 17:09:39