c++ - 使用 auto&& 完美转发返回值

Question

请引用 C++ 模板:完整指南(第 2 版) 中的这句话:

decltype(auto) ret{std::invoke(std::forward<Callable>(op),
                               std::forward<Args>(args)...)};
...
return ret;

Note that declaring ret with auto&& is not correct. As a reference, auto&& extends the lifetime of the returned value until the end of its scope but not beyond the return statement to the caller of the function.

作者说 auto&& 不适合完美转发返回值。但是，decltype(auto) 不是也形成了对 xvalue/lvalue 的引用吗？IMO，decltype(auto) 然后遇到同样的问题。那么，作者的意思是什么？

编辑:

上面的代码片段应该放在这个函数模板中。

template<typename Callable, typename... Args>
decltype(auto) call(Callable&& op, Args&&... args) {
    // here
}

Answer 1

这里有两个推论。一个来自返回表达式，一个来自 std::invoke表达。因为 decltype(auto) is deduced to be the declared type for unparenthesized id-expression , 我们可以关注从 std::invoke 中的推论表达。

引自[dcl.type.auto.deduct] paragraph 5:

If the placeholder is the decltype(auto) type-specifier, T shall be the placeholder alone. The type deduced for T is determined as described in [dcl.type.simple], as though e had been the operand of the decltype.

并引用自[dcl.type.simple] paragraph 4 :

For an expression e, the type denoted by decltype(e) is defined as follows:

• if e is an unparenthesized id-expression naming a structured binding ([dcl.struct.bind]), decltype(e) is the referenced type as given in the specification of the structured binding declaration;

• otherwise, if e is an unparenthesized id-expression or an unparenthesized class member access, decltype(e) is the type of the entity named by e. If there is no such entity, or if e names a set of overloaded functions, the program is ill-formed;

• otherwise, if e is an xvalue, decltype(e) is T&&, where T is the type of e;

• otherwise, if e is an lvalue, decltype(e) is T&, where T is the type of e;

• otherwise, decltype(e) is the type of e.

备注decltype(e)推导为 T而不是 T&&如果e是纯右值。这是与 auto&& 的区别.

所以如果std::invoke(std::forward<Callable>(op), std::forward<Args>(args)...)是纯右值，例如 Callable 的返回类型不是引用，即按值返回，ret被推导为同一类型而不是引用，完美转发了按值返回的语义。