C – 模板中的复制分配操作符

发布时间：2020-12-16 10:01:32 所属栏目：百科来源：网络整理

导读：我正在尝试在模板struct xpair中重载一个复制赋值运算符 template typename First,typename Secondstruct xpair { First first{}; Second second{}; xpair(){} xpair (const First first,const Second second): first(first),second(second) {} xpair operat

我正在尝试在模板struct xpair中重载一个复制赋值运算符

template <typename First,typename Second>
struct xpair {
   First first{};
   Second second{};

   xpair(){}
   xpair (const First& first,const Second& second):
           first(first),second(second) {}


   xpair& operator= (const xpair& that) {
      cout << "*this = " << *this << "      " << "that = " << that << endl;
      cout << "use operator = " << endl;
      *this = that;
      return *this;
   }
};

但是当我用这个代码测试时

using commend = string;
using str_str_pair = xpair<string,string>;
using commend_pair = xpair<commend,str_str_pair>;

commend_pair cmd_pair;
str_str_pair str_pair("a","1");
commend cmd("comment");

cmd_pair.first = cmd;
cmd_pair.second = str_pair;

它给了我无限的输出

use operator = 
*this = {,}      that = {a,1}
use operator = 
*this = {,1}

这是为什么？

解决方法

It gives me infinite output as

Why is that?

因为您已根据自身定义了该函数,请参阅以下代码注释.

xpair& operator= (const xpair& that)
{
    cout << "*this = " << *this << "      " << "that = " << that << endl;
    cout << "use operator = " << endl;

    // Here you're asking for `this` (i.e.,an `xpair` type) to be assigned
    // a `that` (i.e.,another `xpair` type) using the `operator=` which is
    // the function currently being implemented/defined. A function calling
    // itself is recursion and there is no stopping condition so it will
    // continue infinitely.
    *this = that;
    return *this;
}

相反,您的操作应使用该实例的数据成员设置此实例的数据成员.

xpair& operator= (const xpair& that)
{
    cout << "*this = " << *this << "      " << "that = " << that << endl;
    cout << "use operator = " << endl;

    first = that.first;
    second = that.second;
    return *this;
}

（编辑：李大同）

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