为什么C ++模板类型匹配不检索引用限定符'&'? [英] Why C++ template type match doesn't retrieve reference qualifier '&'?

问题描述

我有以下程序:

#include<stdio.h>
template<class T> void f(T t) { 
    t += 1; 
}
template<class T> void g(T &t) {
    t += 10;
}
int main()
{
    int n=0;
    int&i=n;
    f(i);
    g(i);
    printf("%d\n",n);
    return 0;
}

我希望因为 i 是对 n 的引用，所以我希望模板函数 f 应该得到 int&(用于模板类型 T ).但实际上并非如此.程序的输出是 10 ，而不是我期望的 11 .

I expect that because i is a reference to n, so I expect that the template function f should get int& for template type T. But in fact it doesn't. The output of the program is 10, not 11 as I expected.

所以我的问题是，对于 f ，为什么 T 匹配变量<代码>我?这里的规则是什么?

So my question is, for f, why T matches int but not int& of variable i? What's the rule behind here?

谢谢.

推荐答案

除非您使用转发引用，否则模板推导永远不会推导引用类型.因此，您对 f 和 g 的调用都会推断出 T 为 int .

Template deduction never deduces a reference type unless you use a forwarding reference. So your calls to f and g both deduce T as int.

此外，表达式永远不会具有引用类型. i 和 n 是相同的表达式.它们具有类型 int 和值类别 lvalue .

Also, an expression never has reference type. i and n as expressions are identical. They have type int and value category lvalue.

代码 int n = 0;int &i = n; 和 int i = 0; 完全一样int& n = i; ，但 decltype(1).它创建一个具有两个名称的对象，分别是 i 和 n .

The code int n = 0; int &i = n; is exactly the same as int i = 0; int &n = i;, except for decltype(1). It creates one object with two names, i and n.

即使您确实在代码中使用了转发引用，例如 template< class T> void h(T& t)，调用 h(i)和 h(n)会得出同样的方式.

Even if you did use a forwarding reference in your code, e.g. template<class T>void h(T&&t), the calls h(i) and h(n) would deduce the same way.

这种对称性就是为什么您在

This symmetry is why you see many comments on the What is a reference? megathread present references as I have just now, and we consider the description "a reference is a pointer that automatically dereferences" to be misleading.

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