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.