强制编译器选择带有const T&的复制构造函数作为参数 [英] Force compiler to choose copy constructor with const T& as a parameter

问题描述

我正在编写一个具有模板化构造函数和复制构造函数的类。每次我想用非const对象调用复制构造函数时，都会选择模板化构造函数。如何强制编译器选择复制构造函数？

I'm writing a class where I have a templated constructor and copy constructor. Every time I want to call copy constructor with non const object, templated constructor gets chosen. How can I force compiler to choose copy constructor?

这是mcve：

#include <iostream>

struct foo
{
    foo()
    {
        std::cout << "def constructor is invoked\n";
    }

    foo(const foo& other)
    {
        std::cout << "copy constructor is invoked\n";
    }

    template <typename T>
    foo(T&& value)
    {
        std::cout << "templated constructor is invoked\n";
    }
};

int main()
{
    foo first;
    foo second(first);
}

删除功能不是我想要的。

Deleting a function is not what I want.

推荐答案

问题是 first 是可变的，因此对其的引用是 foo& 比 const foo& T& $ c>。

The problem is that first is mutable, so a reference to it is a foo& which binds to the universal reference T&& more readily than const foo&.

想必您打算 T 是任何非foo类吗？

Presumably, you intended that T was any non-foo class?

在这种情况下， enable_if 一点点地表达了编译器的意图，而不必编写大量的伪重载。

In which case a little bit of enable_if chicanery expresses intent to the compiler without having to write a load of spurious overloads.

#include <iostream>

struct foo
{
    foo()
    {
        std::cout << "def constructor is invoked\n";
    }

    foo(const foo& other)
    {
        std::cout << "copy constructor is invoked\n";
    }

    template <typename T, std::enable_if_t<not std::is_base_of<foo, std::decay_t<T>>::value>* = nullptr>
    foo(T&& value)
    {
        std::cout << "templated constructor is invoked\n";
    }

};

int main()
{
    foo first;
    foo second(first);
    foo(6);
}

预期产量：

def constructor is invoked
copy constructor is invoked
templated constructor is invoked

这篇关于强制编译器选择带有const T&的复制构造函数作为参数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！