隐式调用复制构造函数？

Question

我有以下 class，同时定义了普通构造函数和复制构造函数。

#include <iostream>

class Bla
{
public:
    Bla()
    {
        std::cout << "Normal Constructor Called\n";
    }

    Bla(const Bla& other)
    {
        std::cout << "Copy Constructor Called\n";
    }

};

int main() 
{
    Bla a = Bla(); // prints Normal Constructor
}

在主函数中，它按我的预期打印了普通构造函数，并且只打印了普通构造函数。但是，如果我将复制构造函数设为 class 的私有成员，编译器会给出错误

error: ‘Bla::Bla(const Bla&)’ is private within this context

从外观上看，似乎调用了复制构造函数，但我没有看到从中打印出任何内容。复制构造函数是否被隐式调用？这是怎么回事？

Answer 1

在 C++17 之前，复制操作可能是 elided，但复制构造函数仍然需要存在和访问。

This is an optimization: even when it takes place and the copy/move (since C++11) constructor is not called, it still must be present and accessible (as if no optimization happened at all), otherwise the program is ill-formed:

自 C++17 起，由于强制复制省略，不再存在此类问题。

Under the following circumstances, the compilers are required to omit the copy and move construction of class objects, even if the copy/move constructor and the destructor have observable side-effects. The objects are constructed directly into the storage where they would otherwise be copied/moved to. The copy/move constructors need not be present or accessible:

隐式调用复制构造函数？

Copy constructor implicitly called?

c++

constructor

copy-constructor

copy-elision