The documentation says, that size_type
of std::vector
is /usually/ size_t
, which is reasonable, since an implementation can choose to use different.
But why is size_type = size_t
in std::array
. Especially here, as std::array
is used on small µC
a lot, it would be better to let the implemenatation have some freedom.
Is this a doc-defect?
It's defined to be that way because size_t
is defined to be sufficient for all arrays. If you want a smaller type for smaller arrays, you can always narrow when appropriate based on constexpr
values.
template <typename Array>
struct small_array_size
{
using type = size_t
};
template <typename T, size_t N, typename = std::enable_if_t<N < 256>>
struct small_array_size<std::array<T, N>>
{
using type = uint8_t;
};
template <typename T, size_t N, typename = std::enable_if_t<N < 65536>>
struct small_array_size<std::array<T, N>>
{
using type = uint16_t;
};
template <typename Array>
using small_array_size_t = typename small_array_size<Array>::type;
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.