https://gcc.gnu.org/bugzilla/show_bug.cgi?id=78420
Jens Maurer <jens.maurer at gmx dot net> changed:
What |Removed |Added
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CC| |jens.maurer at gmx dot net
--- Comment #16 from Jens Maurer <jens.maurer at gmx dot net> ---
I'd like to point out that there is no prohibition against writing
reinterpret_cast inside a constexpr function. It's just if you call that
function and actually evaluate the reinterpret_cast does the expression turn
into an expression that is not a constant expression.
And there is no requirement that calling a constexpr function with arbitrary
arguments is, in fact, a constant expression in the sense of C++ section 5.20.
There seems to be a tacit understanding that a standard library function marked
as "constexpr" may, in fact, appear in a constant expression if the "obvious"
operations on the arguments are suitable (copy constructor, destructor at
least), but I couldn't find a statement that would make this expectation
explicit.
That means the standard library needs to do its homework to clearly specify
under which circumstances (which argument types) it expects a constexpr
function to be valid in a constant expression. Absent that, simply performing
the reinterpret_cast is the right answer for std::less<T*>, and seems to be
fully conforming with the letter of the current standard.
