在编译时测试endianess：这个constexpr函数是否符合标准？

Question

After some search for a way to check endianess at compile-time I've come up with the following solution:

static const int a{1};

constexpr bool is_big_endian()
{
    return *((char*)&(a)) == 1;
}

GCC accepts this code only in some contexts where constexpr is required:

int b[is_big_endian() ? 12 : 25]; //works
std::array<int, testendian() ? 12 : 25> c;  //fails

For the second case, GCC says error: accessing value of 'a' through a 'char' glvalue in a constant expression . I couldn't find anything in the standard that forbids such thing. Maybe someone could clarify in which case GCC is correct?

Answer 1

This is what I get from Clang 3.1 ToT:

error: constexpr function never produces a constant expression

§5.19 [expr.const]

p1 Certain contexts require expressions that satisfy additional requirements as detailed in this sub-clause; other contexts have different semantics depending on whether or not an expression satisfies these requirements. Expressions that satisfy these requirements are called constant expressions .

p2 A conditional-expression is a core constant expression unless it involves one of the following as a potentially evaluated subexpression:

• [...]
• a reinterpret_cast (5.2.10);

So, (char*)&(a) evaluates to a reinterpret_cast , as such the function is never a valid constexpr function.

Answer 2

You should look into Boost.Detail.Endian

It is a mapping of several architectures to their endianness (through the macros BOOST_BIG_ENDIAN, BOOST_LITTLE_ENDIAN, and BOOST_PDP_ENDIAN). As far as I know, there is no actual way to determine the endianness at compile time, other than a list like this.

For an example implementation that uses Boost.Detail.Endian, you can see the library I'm hoping to get reviewed for submission to Boost: https://bitbucket.org/davidstone/endian/ (the relevant file is byte_order.hpp , but unsigned.hpp is necessary as well if you want to just use my implementation).

Answer 3

如果实现N3620 - 网络字节顺序转换，您将能够使用constexpr ntoh检查字节顺序，但请记住，有一些罕见的体系结构，如中端，您永远无法支持所有这些。