右转转发参考 [英] Rvalue to forwarding references

问题描述

我正在阅读引用折叠规则，我有一个问题:为什么我将右值A 传递给

I'm reading the reference collapsing rules and I have a question: why if I pass a rvalue A to

template<typename T>
void foo(T&&);

T推论为A?

例如如果我将 std :: string()传递给函数 T 推导为 std :: string ，为什么不是 std::string&& ?对我来说，这更有意义，将 T 推导为类型本身的原理是什么?

e.g. if I pass std::string() to the function T is deduced to be std::string, why not std::string&&? It would have made more sense to me, what's the rationale behind deducing T to the type itself?

推荐答案

这仅与我们从模板类型推导中获得的期望基本一致:

This merely lines up with what we expect in general from template type deduction:

template <class T> void vfoo(T );
template <class T> void lfoo(T& );
template <class T> void cfoo(T const& );
template <class T> void ffoo(T&& );

std::string x;
vfoo(x);            // deduce T = std::string
lfoo(x);            // deduce T = std::string
cfoo(x);            // deduce T = std::string
ffoo(x);            // deduce T = std::string& !
ffoo(std::move(x)); // deduce T = std::string

来自原始论文，重点是:

当使用与右值引用匹配的左值参数推导函数模板类型时，该类型被推导为左值引用类型.当给定推导一个右值参数时，类型推导与其他类型一样进行.

这是左值扣除案例，这是例外，这就是为什么它在类型推断规则中得到一个额外的句子.右值情况很典型-它与粘贴推论类型的简单思维模型一致，以查看最终要使用的功能.用 std :: string 称为 T& ?获取 T = std :: string ，以便该参数读取 std :: string&&.查看.

It's the lvalue deduction case that's the exceptional one, which is why it gets an extra sentence in the type deduction rules. The rvalue case is typical - it lines up with the simple mental model of pasting in the deduced types to see what function you end up with. Called T&& with a std::string? Get T = std::string so that the argument reads std::string&&. Check.

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