当等效函数只调用一次时，为什么转换运算符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，并定义了一个等效的转换函数。

#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);
}

我没有优化编译时的输出是

-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，但在它调用函数convert() 后只调用一次，该函数执行相同的操作。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() 不会导致生成该对象。为什么调用转换运算符后程序调用拷贝构造函数的次数和调用convert()后程序调用拷贝构造函数的次数会有差异？

Answer 1

定义

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

真的

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。