为什么派生类不使用基类operator =(赋值运算符)? [英] Why doesn't a derived class use the base class operator= (assignment operator)?

问题描述

以下是实际问题的简化版本.该代码似乎没有生成Derived临时对象，而是对其进行复制，而不是调用Base::operator=(int).由于函数签名似乎完美匹配，为什么不使用基本赋值运算符?这个简化的示例不会显示任何不良影响，但是原始代码在析构函数中具有副作用，会造成各种破坏.

Following is a simplified version of an actual problem. Rather than call Base::operator=(int), the code appears to generate a temporary Derived object and copy that instead. Why doesn't the base assignment operator get used, since the function signature seems to match perfectly? This simplified example doesn't display any ill effects, but the original code has a side-effect in the destructor that causes all kinds of havoc.

#include <iostream>
using namespace std;

class Base
{
public:
   Base()
   {
      cout << "Base()\n";
   }

   Base(int)
   {
      cout << "Base(int)\n";
   }

   ~Base()
   {
      cout << "~Base()\n";
   }

   Base& operator=(int)
   {
      cout << "Base::operator=(int)\n";
      return *this;
   }
};

class Derived : public Base
{
public:
   Derived()
   {
      cout << "Derived()\n";
   }

   explicit Derived(int n) : Base(n)
   {
      cout << "Derived(int)\n";
   }

   ~Derived()
   {
      cout << "~Derived()\n";
   }
};

class Holder
{
public:
   Holder(int n)
   {
      member = n;
   }

   Derived member;
};

int main(int argc, char* argv[])
{
   cout << "Start\n";
   Holder obj(1);
   cout << "Finish\n";

   return 0;
}

输出为:

Start
Base()
Derived()
Base(int)
Derived(int)
~Derived()
~Base()
Finish
~Derived()
~Base()

http://ideone.com/TAR2S

推荐答案

这是编译器生成的operator=方法与

This is a subtle interaction between a compiler-generated operator= method and member function hiding. Since the Derived class did not declare any operator= members, one was implicitly generated by the compiler: Derived& operator=(const Derived& source). This operator= hid the operator= in the base class so it couldn't be used. The compiler was still able to complete the assignment by creating a temporary object using the Derived(int) constructor and copy it with the implicitly generated assignment operator.

因为执行隐藏的功能是隐式生成的，并且不是源代码的一部分，所以很难发现它.

Because the function doing the hiding was generated implicitly and wasn't part of the source, it was very hard to spot.

这可能是通过使用int构造函数上的explicit关键字发现的-编译器将发出错误，而不是自动生成临时对象.在原始代码中，隐式转换是一种常用的功能，因此未使用explicit.

This could have been discovered by using the explicit keyword on the int constructor - the compiler would have issued an error instead of generating the temporary object automatically. In the original code the implicit conversion is a well-used feature, so explicit wasn't used.

解决方案非常简单，Derived类可以从Base类显式提取定义:

The solution is fairly simple, the Derived class can explicitly pull in the definition from the Base class:

using Base::operator=;

http://ideone.com/6nWmx

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