强制在赋值运算符中使用隐式强制转换

Question

在下面的代码中，有什么方法可以强制编译器使用隐式转换为 bool（在 b3 = b2 赋值中）而不是生成复制赋值运算符 而不使用转换运算符 ？不幸的是，在显式删除复制分配代码后不会构建。缺少显式转换对我的框架很重要。

class Base
{
public :

   Base& operator=(const Base&) = delete;
   virtual Base& operator=(bool state) { state_ = state; }
   virtual operator bool() const { return state_; }
   virtual bool operator!() const { return !state_; }

protected :
   bool state_;
};

int main(void)
{
   Base b1, b2, b3;
   b1 = true;
   b2 = !b1;
   b3 = b2;

   return 0;
}

更新：错误是

test.cpp: In function ‘int main()’:
test.cpp:20:9: error: use of deleted function ‘Base& Base::operator=(const Base&)’
    b3 = b2;
         ^~
test.cpp:6:10: note: declared here
    Base& operator=(const Base&) = delete;
          ^~~~~~~~

更新2：正如@serge-ballesta 所说，基本赋值运算符就足够了

#include <iostream>

class Base
{
public :

   virtual Base& operator=(const Base &rhs) { std::cout << "BASE = BASE" << std::endl; return *this = static_cast<bool>(rhs); };
   virtual Base& operator=(bool state) { std::cout << "BASE = bool" << std::endl;  state_ = state; return *this; }
   virtual operator bool() const { return state_; }
   virtual bool operator!() const { return !state_; }

protected :
   bool state_;
};

class Derived :
   public Base
{
public :
   virtual Base& operator=(bool state) { std::cout << "DERIVED = bool" << std::endl; state_ = state; /* And something more */ return *this; }
};

int main(void)
{
   Base b1, b2, b3;
   b1 = true;
   b2 = !b1;
   b3 = b2;

   Derived d1, d2, d3, d4;
   d1 = true;
   d2 = !d1;
   d3 = d2;
   d4 = b3;

   return 0;
}

输出为：

BASE = bool      # b1 = true;
BASE = bool      # b2 = !b1;

BASE = BASE      # b3 = b2;
BASE = bool      # ditto

DERIVED = bool   # d1 = true;
DERIVED = bool   # d2 = !d1;

BASE = BASE      # d3 = d2;
DERIVED = bool   # ditto

DERIVED = bool   # d4 = b3;

有趣的是，在最后一个案例中，隐式转换是按照我的意愿完成的。

Answer 1

明确删除赋值运算符意味着您希望 class 不可复制。在这里，您只希望复制赋值使用 bool 转换：

Base& operator=(const Base& other) {
    *this = static_cast<bool>(other);
    return *this;
}

不幸的是，您将不得不覆盖派生的 classes 中的赋值运算符以强制使用这个：

class Derived: public Base {
public:
    Derived& operator=(Derived& other) {
        Base::operator = (other);
        return *this;
    }   
    ...
};

强制在赋值运算符中使用隐式强制转换

Force using implicit cast in assign opertor

c++

operator-overloading