模板构造函数优先于普通复制和移动构造函数？ [英] Template Constructor Taking Precedence Over Normal Copy and Move Constructor?

问题描述

以下程序的输出...

The output of the following program...

#include <iostream>

using namespace std;

struct X
{
    X(const X&)              { cout << "copy" << endl; }
    X(X&&)                   { cout << "move" << endl; }

    template<class T> X(T&&) { cout << "tmpl" << endl; }
};

int main()
{
    X x1 = 42;
    X x2(x1);
}

是

tmpl
tmpl

所需的输出为：

tmpl
copy

为什么具体的复制构造函数不优先于模板构造函数？

Why doesn't the concrete copy constructor take precedence over the template constructor?

总有办法修复它，以便复制并移动

Is there anyway to fix it so that the copy and move constructor overloads will take precedence over the template constructor?

推荐答案

选择构造函数时，正常的重载解析规则仍然适用-构造函数采用与采用const左值引用的构造函数相比，非const左值引用（对于在扣除参数后的模板构造函数而言）是更好的匹配。

Normal overload resolution rules still apply when choosing the constructor - and a constructor taking a non-const lvalue reference (for the template constructor after argument deduction) is a better match than a constructor taking a const lvalue reference.

当然，您可以添加另一个采用非常量左值引用的重载，即

You could, of course, just add another overload taking a non-const lvalue reference, i.e.

X(X&)              { cout << "copy" << endl; }

更新：模板构造函数更匹配的其他情况：

Update: Other cases where the template constructor is a better match:

const X f()
{ return X(); }

struct Y : X
{ Y() { } };

int main()
{
  X x3(f()); // const-qualified rvalue
  Y y;
  X x4(y); // derived class
}

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