c++ - SFINAE and decltype(auto)

Question

If a function template returns decltype(auto) (or another type specifier using auto) but the return statement would be ill-formed, does SFINAE result? Is the return statement considered to be the immediate context of the function signature?

Nothing in the N3690 draft seems to require this. By default, I guess SFINAE does not apply.

This seems unfortunate because you can write a function to forward to another function, but you cannot make its existence conditional on the delegate as when writing longhand. Furthermore, checking the existence of a peer nonstatic member function cannot be done without decltype(auto) because this cannot be used in a function signature. However this indicates a fundamental problem, as decltype(auto) provides a path to considering the class type as complete within a member signature, where it's not.

Has a proposal been written, or has the problem been formally analyzed anywhere?

The ability to treat the class type as complete within a member signature may have other implications… but that's just fodder for another question.

Answer 1

but the return statement would be ill-formed, does SFINAE result?

The proposal-n3638 says,

SFINAE

Since the return type is deduced by instantiating the template, if the instantiation is ill-formed, this causes an error rather than a substitution failure. This allows an auto function to return a lambda, which is not possible using the decltype(returned expression) pattern.

Hope that is what you're looking for.