标准如何规定下面代码中对A :: operator int（）的调用？

Question

根据程序中的注释，在下面的代码中调用

A::operator int()。我想我很清楚这里发生了什么。但我想更准确地了解标准如何支持这些电话？我知道这种（隐式）运算符是在复制初始化中调用的，这似乎不是这里的情况。

#include <iostream>

class A {
    int i;
public:
    A(int j) : i(j){ std::cout << "constructor\n";  }
    A& operator=(int j) { std::cout << "operator =\n"; i = j; return *this; }
    A& operator+=(const A& rhs) { i += rhs.i; return *this; }
    const A operator+(const A& rhs) const { return A(*this) += rhs; }
    A& operator-=(const A& rhs) { i -= rhs.i; return *this; }
    const A operator-(const A& rhs) const { return A(*this) -= rhs; }
    operator int() const { std::cout << "operator int()\n"; return i; }
};

int main()
{
    A a(1);                 //  Invokes constructor A(int), printing "constructor"
    A b(2);                 //  Idem
    A c = a + b;            //  Invokes a.operator+(b) followed by a call to the default copy constructor which copies
                            //  the object referenced by the return of a + b into c.
    std::cout << c << '\n'; //  operator int() is called to convert the object c into an int, printing "operator int()"
                            //  followed by the number 3.
    c = a - b;              //  Invokes a.operator-(b) followed by a call to the default assignment operator which
                            //  copies the object referenced by the return of a - b into c.

    std::cout << c << '\n'; //  operator int() is called to convert the object c into an int, printing "operator int()"
                            //  followed by the number -1.

    c = (a - b) * c;        //  Invokes a.operator-(b) followed by two calls to operator int(), one to convert the
                            //  result of a - b into an int and another to convert c into an int. Finally the special
                            //  assignment operator, operator=(int) is called to assign the int resultant from the
                            //  expression (a - b) * c to the object c, printing "operator =".
}