c ++：cast操作符与赋值运算符与转换构造函数优先级 [英] c++: cast operator vs. assign operator vs. conversion constructor priority

问题描述

让我们有这个代码：

  Test1 t1; 
 Test2 t2; 
 t1 = t2; 
  

我相信有三种（或更多）方法如何实现 t1 =

• 重载赋值操作符 Test1


• 可以在 Test2中重载类型转换运算符 Test1（const Test2&）转换构造函数

根据我的GCC测试，

转换构造函数和类型转换运算符（模糊）

>

const转换构造函数和const类型转换运算符（模糊）

请帮助我理解为什么会有这个优先级。

我使用此代码进行测试（取消注释一些行以尝试）

  struct Test2; 
 struct Test1 {
 Test1（）{} 
 Test1（const Test2& t）{puts（const constructor wins）; } 
 // Test1（Test2& t）{puts（constructor wins）; } 
 // Test1& operator =（Test2& t）{puts（assign wins）; } 
}; 
 
 struct Test2 {
 Test2（）{} 
 //操作符Test1（）const {puts（const cast wins）; return Test1（）; } 
 // operator Test1（）{puts（cast wins）; return Test1（）; } 
}; 
 
 
 int main（）{
 Test1 t1; 
 Test2 t2; 
 t1 = t2; 
 return 0; 
} 
  

解决方案

c> t1 = t2; 等效于：

  t1.operator = 
  

现在通常的重载解决规则适用。如果有直接匹配，那就是所选的匹配。如果不是，那么隐式转换被认为用于（自动生成的，隐式定义）拷贝赋值运算符。

有两种可能的隐式，用户定义转换。所有用户定义的转化计数相等，如果两者都定义，则重载是不明确的：




• 转换 Test1 :: Test1（Test2 const&）到




• 将 t2 转换为 Test1 通过 Test2 :: operator Test1（）const cast运算符。

Let's have this code:

Test1 t1;
Test2 t2;
t1 = t2;

I believe there are three (or more?) ways how to implement t1 = t2

• to overload assign operator in Test1
• to overload type cast operator in Test2
• to create Test1(const Test2&) conversion constructor

According to my GCC testing, this is the priority of what is used:

1. assign operator
2. conversion constructor and type cast operator (ambiguous)
3. const conversion constructor and const type cast operator (ambiguous)

Please help me understand why this priority.

I use this code for testing (uncomment some lines to try out)

struct Test2;
struct Test1 {
  Test1() { }
  Test1(const Test2& t) { puts("const constructor wins"); }
//  Test1(Test2& t) { puts("constructor wins"); }
//  Test1& operator=(Test2& t) { puts("assign wins"); }
};

struct Test2 {
  Test2() { }
//  operator Test1() const { puts("const cast wins"); return Test1(); }
//  operator Test1() { puts("cast wins"); return Test1(); }
};


int main() {
  Test1 t1;
  Test2 t2;
  t1 = t2;
  return 0;
}

解决方案

The statement t1 = t2; is equivalent to:

t1.operator=(t2);

Now the usual rules of overload resolution apply. If there's a direct match, that's the chosen one. If not, then implicit conversions are considered for use with the (automatically generated, "implicitly defined") copy-assignment operator.

There are two possible implicit, user-defined conversions. All user-defined conversions count equal, and if both are defined, the overload is ambiguous:

• Convert t2 to a Test1 via the Test1::Test1(Test2 const &) conversion constructor.

• Convert t2 to a Test1 via the Test2::operator Test1() const cast operator.

这篇关于c ++：cast操作符与赋值运算符与转换构造函数优先级的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！