SFINAE和decltype（auto） [英] SFINAE and decltype(auto)
问题描述
如果函数模板返回 decltype（auto）（或使用 auto 的另一个类型说明符）会不会形成，SFINAE结果吗？是 return 语句被认为是函数签名的直接上下文吗？
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?
N3690草案中没有什么似乎需要这个。默认情况下，我猜SFINAE不适用。
Nothing in the N3690 draft seems to require this. By default, I guess SFINAE does not apply.
这似乎不幸，因为你可以写一个函数转发到另一个函数，但你不能让它的存在条件的代理当写长文时。此外，无法使用 decltype（auto）检查对等非静态成员函数的存在，因为不能使用 this 在函数签名中。然而，这表明一个根本问题，因为 decltype（auto）提供了一个路径来考虑类类型在成员签名中是完整的，而不是成员签名。
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.
是否已写入提案，或者已在任何地方正式分析问题？
Has a proposal been written, or has the problem been formally analyzed anywhere?
能够将类类型视为完整的成员签名可能有其他影响...但是这只是另一个问题的饲料。
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.
推荐答案
语句将是不成形的，SFINAE是否结果？
but the return statement would be ill-formed, does SFINAE result?
SFINAE
由于通过实例化模板来推导返回类型，如果实例化不成功，则会导致错误，而不是替换失败。这允许一个自动函数返回一个lambda，这是不可能使用decltype（返回的表达式）模式。
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.
希望这是什么您正在寻找。
Hope that is what you're looking for.
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