模板赋值运算符重载之谜 [英] Template assignment operator overloading mystery

问题描述

我有一个简单的结构 Wrapper，以两个模板化的赋值运算符重载来区分:

I have a simple struct Wrapper, distinguished by two templated assignment operator overloads:

template<typename T>
struct Wrapper {

  Wrapper() {}

  template <typename U>
  Wrapper &operator=(const Wrapper<U> &rhs) {
    cout << "1" << endl;
    return *this;
  }
  template <typename U>
  Wrapper &operator=(Wrapper<U> &rhs) {
    cout << "2" << endl;
    return *this;
  }
};

然后我声明 a 和 b:

I then declare a and b:

Wrapper<float> a, b;
a = b;

将 b 赋值给 a 将使用上面的非常量模板赋值运算符重载，并显示数字2".

assigning b to a will use the non-const templated assignment operator overload from above, and the number "2" is displayed.

令我困惑的是:如果我声明 c 和 d，

What puzzles me is this: If I declare c and d,

Wrapper<float> c;
const Wrapper<float> d;
c = d;

并将d赋值给c，两个赋值运算符重载都没有使用，不显示输出；所以调用了默认的复制赋值运算符.为什么将 d 分配给 c 不使用提供的 const 重载赋值运算符?或者相反，为什么将 b 分配给 a 不 使用默认的复制赋值运算符?

and assign d to c, neither of the two assignment operator overloads is used, and no output is displayed; so the default copy assignment operator is invoked. Why does assigning d to c not use the const overloaded assignment operator provided? Or instead, why does assigning b to a not use the default copy assignment operator?

推荐答案

为什么将 d 赋值给 c 不使用提供的 const 重载赋值运算符?

Why does assigning d to c not use the const overloaded assignment operator provided?

隐式声明的复制赋值运算符，声明如下，仍然生成:

The implicitly-declared copy assignment operator, which is declared as follows, is still generated:

Wrapper& operator=(const Wrapper&);

运算符模板不会抑制隐式声明的复制赋值运算符的生成.由于参数(const 限定的 Wrapper)与此运算符的参数(const Wrapper&)完全匹配，因此在重载解析期间选择它.

An operator template does not suppress generation of the implicitly-declared copy assignment operator. Since the argument (a const-qualified Wrapper) is an exact match for the parameter of this operator (const Wrapper&), it is selected during overload resolution.

未选择操作符模板并且没有歧义，因为——所有其他条件都相同——在重载解析期间，非模板比模板更匹配.

The operator template is not selected and there is no ambiguity because--all other things being equal--a nontemplate is a better match during overload resolution than a template.

为什么将 b 赋值给 a 不使用默认的复制赋值运算符?

Why does assigning b to a not use the default copy assignment operator?

参数(非const限定的Wrapper)与采用Wrapper<U>& 的操作符模板相比，比隐式-声明的复制赋值运算符(它需要一个 const Wrapper&.

The argument (a non-const-qualified Wrapper) is a better match for the operator template that takes a Wrapper<U>& than for the implicitly-declared copy assignment operator (which takes a const Wrapper<U>&.

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