c++ - 为什么转换运算符在等效函数只调用一次时调用复制构造函数两次？

Question

我在 Visual Studio 2017 中运行。我的程序定义了一个带有转换运算符的类，并定义了等效的转换函数。

#include <iostream>
#include <string>

template<class T, class U>
struct A
{
    A();
   ~A() { std::cout << "In A destructor\n"; delete n; }
    A(T i);
    A(const A& a);
    operator A<U,T>();
    T* n;
};

template<class T, class U>
A<T,U>::A()
{
    n = new T;
   *n = 0;
}

template<class T, class U>
A<T,U>::A(T i)
{
    n = new T;
   *n = i;
}

template<class T, class U>
A<T,U>::A(const A& a)
{
    std::cout << "In A copy constructor\n";
    n = new T;
   *n = *(a.n);
}

template<class T, class U>
A<T,U>::operator A<U,T>()
{
    A<U,T> aut;
    *(aut.n) = static_cast<U>(*n);
    return aut;
}

template<class T, class U>
std::ostream& operator<<(std::ostream& os, A<T,U>& a)
{
    os << *(a.n);
    return os;
}

template<class T, class U>
A<U,T> convert(A<T,U>& a)
{
    A<U,T> c;
   *(c.n) = static_cast<U>(*(a.n));
    return c;
}

int main()
{
    std::string s;

    A<int, unsigned int> a1(-1);
    std::cout << a1 << "\n";

    A<unsigned int, int> a2 = A<unsigned int, int>(a1);
    std::cout << a2 << "\n";

    A<unsigned int, int> a3 = convert(a1);
    std::cout << a3 << "\n";

    std::cout << "Press ENTER to exit\n";
    getline(std::cin, s);
}

我在没有优化的情况下编译时的输出是

-1
In A copy constructor
In A destructor
In A copy constructor
In A destructor
4294967295
In A copy constructor
In A destructor
4294967295
Press ENTER to exit

复制构造函数在程序调用转换运算符后被调用两次，但在调用函数 convert() 后只调用一次，后者执行相同的操作。似乎在对转换运算符的调用中，程序正在创建一个临时对象，函数 convert() 不会导致创建该对象。为什么程序在调用转换运算符后调用复制构造函数的次数与程序在调用convert()后调用复制构造函数的次数不同？

Answer 1

定义

A<unsigned int, int> a2 = A<unsigned int, int>(a1);

真的是

A<unsigned int, int> a2 = A<unsigned int, int>(a1.operator A<unsigned int, int>());

相当于

A<unsigned int, int> temporary_compiler_generated_object = a1.operator A<unsigned int, int>();
A<unsigned int, int> a2 = temporary_compiler_generated_object;

这里需要调用临时对象的复制构造函数以及a2。