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SFINAE and decltype(auto)

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.

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Potatoswatter Avatar asked Jul 12 '13 06:07

Potatoswatter


People also ask

What is the difference between auto and decltype?

'auto' lets you declare a variable with a particular type whereas decltype lets you extract the type from the variable so decltype is sort of an operator that evaluates the type of passed expression.

What is decltype used for?

The decltype type specifier yields the type of a specified expression. The decltype type specifier, together with the auto keyword, is useful primarily to developers who write template libraries. Use auto and decltype to declare a template function whose return type depends on the types of its template arguments.


2 Answers

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.

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Nawaz Avatar answered Oct 21 '22 04:10

Nawaz


Following up on Nawaz's link, the remaining questions are answered by N3690 §7.1.6.4/11:

If the type of an entity with an undeduced placeholder type is needed to determine the type of an expression, the program is ill-formed.

This means that even if SFINAE worked with return type deduction, it couldn't be used to query one function declaration from another. The signature is essentially invalid until the return statement is processed, which occurs at the closing brace of the class {} definition, and after the definitions of preceding members have been processed.

In a sense, all member decltype(auto) functions are incomplete with respect to preceding functions in the same class:

struct s {
    void f() { a(); } // error: use of ‘auto s::a()’ before deduction of ‘auto’
    auto a() { return 3; }
};

This is GCC's complaint; it goes away if the member declarations are reversed. This is because the function definitions are processed in order of declaration, when the } from the class definition is reached. If the statement a(); is processed before the return 3;, then the program is ill-formed.

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Potatoswatter Avatar answered Oct 21 '22 03:10

Potatoswatter