引用模板作为参数的模板参数推导 [英] Template argument deduction for references as arguments

问题描述

我试图深刻理解模板参数推导。
我不了解的一点是，如何应用标准
中的规则

解决方案

A 是表达式的类型。表达式类型由 [expr.type] / 1 描述：

如果表达式最初的类型为对T的引用（[dcl.ref]，[dcl.init.ref]），则类型调整为T。

所以这里 A 是 int 。

此表达式是 lvalue ，但不会播放角色，因为 P 不是参考。

I am trying to profoundly understand Template Argument Deduction. One point I am not understanding is, how I should apply the rules in the standard here for the types A and P for the following case (there is sadly no example on cppreference.com, see below the relevant section)

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

void call_with_reference(int& r) {
    foo(r)
}

• P is no reference typ:
  which gives P := T
• A := int&

-> Match P and A which gives: T is deduced to int&

which is cleary wrong. Where is the rule in the standard that says references from A are removed? A non-confusing, unambiguous clear answer would be very much appreciated.

Relevant Section:

解决方案

A is the type of an expression. Expression type is described by [expr.type]/1:

If an expression initially has the type "reference to T" ([dcl.ref], [dcl.init.ref]), the type is adjusted to T.

So here A is int.

This expression is an lvalue but that will not play any role since P is not a reference.

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