模板分配运算符重载奥秘 [英] Template assignment operator overloading mystery

问题描述

我有一个简单的结构 Wrapper ，通过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：

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

指派 b 至 a

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

对我有用[0]丢个板砖[0]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 分配给 c ，则不使用两个赋值运算符重载，并且不显示任何输出;因此将调用默认的副本分配操作符。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，为什么将 b 分配给 a 不是使用默认的副本赋值运算符？

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

为什么指派 d 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?

隐式声明仍然生成如下声明的复制赋值运算符：

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

Wrapper& operator=(const Wrapper&);

运算符模板不会禁止生成隐式声明的复制赋值运算符。因为参数（一个const限定的 Wrapper ）是这个运算符的参数的完全匹配（ const Wrapper& ），在过载分辨率期间进行选择。

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.

没有选择运算符模板，并且没有歧义，因为 - 所有其他事情都相等 - 在重载解析期间，非模板是比模板更好的匹配。 / p>

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 指定为 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 ）比对于隐含声明的拷贝赋值运算符（其采用了 Wrapper< U>& a const Wrapper< U>& 。

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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