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Subject: Re: Why is size_type in std::array an alias for size_t?
From: Peter C++ (peter.cpp_at_[hidden])
Date: 2019-11-09 01:27:34


If you know that your size_type values always fit into an uint_8t you can store it there if needed, eg, to iterate. Std::array never keeps its size somewhere at run time, so I fail to see the issue.

sent from a mobile device so please excuse strange words due to autocorrection.
Prof. Peter Sommerlad
peter.Sommerlad_at_[hidden]
+41-79-432 23 32

> On 9 Nov 2019, at 07:24, Wilhelm Meier via Std-Discussion <std-discussion_at_[hidden]> wrote:
>
> 
>
>> Am 09.11.19 um 01:58 schrieb Thiago Macieira via Std-Discussion:
>>> On Thursday, 7 November 2019 22:46:22 PST Wilhelm Meier via Std-Discussion
>>> wrote:
>>> I wonder, if this is intentional, since std::array could declare the
>>> size_type depending on the actual size of the array, e.g. an
>>> implementation could declare size_type an alias to uint8_t if the size
>>> of the array is less than 256 or uint16_t if the size of the array is
>>> less than 65536.
>>
>> No, this is intentional. The std::array::size_type should be the type that
>> sizeof(actual_array) returns. That's size_t.
>>
>
> Well, on one side this is understandable, but on the other side an
> implementation für small µC could get some benefit from declaring
> size_type to the small unsigned integer possible for the actual size of
> the array. And I don't see any obstackles doing so.
>
>
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