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From: "daniel.kruegler at googlemail dot com" <gcc-bugzilla@gcc.gnu.org>
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Subject: [Bug c++/53473] [C++11] static constexpr noexcept cannot be specialized
Date: Fri, 25 May 2012 09:24:00 -0000
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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D53473

--- Comment #3 from Daniel Kr=C3=BCgler <daniel.kruegler at googlemail dot =
com> 2012-05-25 09:07:20 UTC ---
(In reply to comment #2)
> Does the standard allow exceptions in constexpr? A throw is not exactly a
> return statement, but according to the rule "constexpr function shall sat=
isfy
> [...] exactly one return statement" I'd expect a constexpr function can n=
ever
> throw anyway. Thus the noexcept itself seems to make no sense in the first
> place.

No, both concepts of constant expressions and exception-specifications are
independent decisions. The following is a perfectly valid constexpr functio=
n:

#include <stdexcept>

constexpr int validating_abs(int val) noexcept(false)
{ return val < 0 ? throw std::runtime_error("negative") : val; }

int main()
{
  constexpr int v1 =3D validating_abs(1); // OK
  constexpr int v2 =3D validating_abs(-1); // Error
  int v =3D -1;
  try {
    int v3 =3D validating_abs(v); // OK, runtime validation
  } catch (std::runtime_error&) {}
}

It is just a fact, that the "effective expression" is relevant when we cons=
ider
constant expressions and throw expressions, like for v2. But constexpr
functions can also be called in non-constant contexts - like for v3 - where
this restriction does not exist.