如何remove_reference禁用模板参数推导？ [英] How remove_reference disable template argument deductions?

问题描述

根据此链接，禁止模板参数推导 std :: forward 和 std :: remove_reference 的帮助我们实现了这一目标。但是使用 remove_reference 如何防止模板推导在这里发生？

According to this link, template argument deduction disallowed for std::forward and std::remove_reference is helping us to achieve that. But how does using remove_reference prevent template deduction from happening here?

template <class S>
S&& forward(typename std::remove_reference<S>::type& t) noexcept
{
    return static_cast<S&&>(t);
}

推荐答案

表达式 typename std :: remove_reference< S> :: type 中的S 是非推断上下文（具体来说因为 S 出现在嵌套名称说明符中，该类型使用 qualified-id 指定）。顾名思义，非推论上下文是无法推导模板参数的上下文。

S in the expression typename std::remove_reference<S>::type is a non-deduced context (specifically because S appears in the nested-name-specifier of a type specified using a qualified-id). Non-deduced contexts are, as the name suggests, contexts in which the template argument cannot be deduced.

这种情况提供了一个简单的示例来理解原因。假设我有：

This case provides an easy example to understand why. Say I had:

int i;
forward(i);

S 是什么？可以是 int ， int& 或 int& -所有这些类型都将为函数产生正确的参数类型。编译器根本无法确定您真正在这里表示的哪个 S -因此不会尝试。这是不可推论的，因此您必须明确提供您指的是 S ：

What would S be? It could be int, int&, or int&& - all of those types would yield the correct argument type for the function. It's simply impossible for the compiler to determine which S you really mean here - so it doesn't try. It's non-deducible, so you have to explicitly provide which S you mean:

forward<int&>(i); // oh, got it, you meant S=int&

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