Trailing return-type syntax fails with noexcept specifier?

Question

This code work as expected:

void f() noexcept {}

But the following fails with an error in GCC 4.7.2:

auto f() -> void noexcept {}

// error: expected initializer before ‘noexcept’

Articles I've read haven't said anything about not being able to specify noexcept in a training return-type. Is this a bug (and had it been fixed in the newest version of GCC)? Or is this prohibited explicitly by the Standard?

Answer 1

That is not the correct syntax. It should be:

auto f() noexcept -> void { }

Per paragraph 8.4.1/2 of the C++11 Standard:

D1 ( parameter-declaration-clause ) cv-qualifier-seq(opt)

ref-qualifier(opt) *exception-specification(opt)* attribute-specifier-seq(opt) *trailing-return-type(opt)*

as described in 8.3.5. A function shall be defined only in namespace or class scope.