为什么需要公共副本构造函数，即使它没有被调用？ [英] Why is a public copy constructor required even if it is not invoked?

问题描述

拥有公共副本构造函数将使小程序
编译，但不显示副作用复制。

Having a public copy constructor will make the little program compile, but not showing the side effect "Copy".

#include <iostream>

class X
{
    public:
    X(int) { std::cout << "Construct" << std::endl; }

    // Having a public copy constructor will make the little program 
    // compile, but not showing the side effect "Copy".

    private:
    X(const X&) { std::cout << "Copy" << std::endl; }

    private:
    X& operator = (const X&);
};

int main() {
    X x = 1;
    return 0;
}

推荐答案

所涉及的C ++标准：

Here are the relevant bits of the C++ standard that are involved:

[dcl.init] / 16，bullet 6，子项目1： - 初始化，或者如果是复制初始化，其中源类型的cv非限定版本与目标类的类是相同的或是派生类，则构造函数被考虑.... 如果没有

在其他方面字，无论编译器优化是否可以省略副本，初始化都是错误的，因为没有适用的构造函数。当然，一旦你复制公开，以下部分适用：

In other words, it doesn't matter if a compiler optimization could elide the copy, the initialization is ill-formed because there are no applicable constructors. Of course, once you make the copy constuctor public, the following section applies:

[class.copy] / 31：当某些标准met，允许实现省略类对象的复制/移动构造，即使对于对象的复制/移动构造函数和/或析构函数具有副作用....这种复制/移动操作的检测，称为复制检测，在以下情况下允许（可以组合使用以消除多个副本）：

[class.copy]/31: When certain criteria are met, an implementation is allowed to omit the copy/move construction of a class object, even if the copy/move constructor and/or destructor for the object have side effects.... This elision of copy/move operations, called copy elision, is permitted in the following circumstances (which may be combined to eliminate multiple copies):

项目符号3：当未绑定到引用的临时类对象12.2）将被复制/移动到具有相同的cv非限定类型的类对象，通过将临时对象直接构造到省略的复制/移动的目标中，可以省略复制/移动操作

bullet 3: when a temporary class object that has not been bound to a reference (12.2) would be copied/moved to a class object with the same cv-unqualified type, the copy/move operation can be omitted by constructing the temporary object directly into the target of the omitted copy/move

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