在C ++ 11或更高版本中有没有办法让constexpr确定没有UB的字节序？

Question

Note that earlier similar questions I found were before C++11 and/or included UB and/or could not be a constexpr.

Not a dup of Is there a way to do a C++ style compile-time assertion to determine machine's endianness? or similar

AFAIK there is little endian , big endian , and other . At a minimum I need to at least not compile if other , even better if other architectures can be added

constexpr and endianness was asked earlier and does not include other , which would leave other architectures ill defined as one or the other

Basically I want to be able to specialize a template based on the target architecture's endianness

Answer 1

As part of my hash_append work I hope to provide what you're asking for:

https://github.com/HowardHinnant/hash_append/blob/master/endian.h

other would be detected by:

endian::native != endian::little && endian::native != endian::big

The first static_assert in this header is currently incorrect with respect to the other issue and should be removed.

This header is very easy to provide for any given platform. But of course it is not portable, and thus is ideal to have it be provided by your std::lib implementor instead.