重载运算符需注意:
1.重载=运算符时容易忘记写返回值。
2.重载赋值运算符时，记得加const，因为赋值操作必须是固定的右值。
3重载时，写在类中的只能有一个参数(实际有两个参数，另外一个是this指针，我们看不见而已)，需要两个参数的时候，要写在类外，用友元在类内声明。
4.重载递增运算符时，要注意哪个要加引用，哪个不用加引用。
//重载前置++运算符 返回引用是为了一直对一个数据进行递增操作MyInteger &operator++() {//先进行++运算m_Num++;//再将自身做一个返回return *this;}//重载后置++运算符//这个int代表的是占位参数，可以用于区分前置和后置递增MyInteger operator++(int) { //后置递增返回的是值，不能是引用//先 返回结果MyInteger temp = *this;//后 递增m_Num++;//最后将记录结果做返回return temp;
代码如下:
#include <iostream>
using namespace std;class Base
{
public:Base():num(0){}Base(const Base &a){this->num = a.num;int len = strlen(a.name);this->name = new char[len + 1];for (int i = 0; i < len; i++) this->name[i] = a.name[i];this->name[len] = '\0';}~Base(){if (name != nullptr) delete[] name;}Base(const char *name, int num) :num(num){int len = strlen(name);this->name = new char[len+1];for (int i = 0; i < len; i++) this->name[i] = name[i];this->name[len] = '\0';}Base operator+(Base &a){Base tmp;tmp.num = this->num + a.num;int len1 = strlen(this->name);int len2 = strlen(a.name);int len3 = len1 + len2;tmp.name = new char[len3 + 1];for (int i = 0; i < len1; i++) tmp.name[i] = this->name[i];for (int i = len1; i < len3; i++) tmp.name[i] =a.name[i-len1];tmp.name[len3] = '\0';return tmp;}Base operator+( const int a){Base tmp(*this);tmp.num += a;return tmp;}friend Base operator+(const int, Base&);friend ostream&operator<<(ostream &, const Base&);friend istream &operator>>(istream &, Base&);Base operator()(const char a[], int b){if (name != nullptr) delete[] name;int len = strlen(a);name = new char[len + 1];for (int i = 0; i < len; i++) name[i] = a[i];name[len] = '\0';num = b;return *this;}bool operator==(const Base &a){if (name == a.name && num == a.num) return true;int len1 = strlen(a.name);int len2 = strlen(this->name);if (len1 == len2 && num == a.num){for (int i = 0; i < len1; i++){if (name[i] != a.name[i]) return false;}return true;}return false;}bool operator !=(const Base &a){if (*this == a) return false;else return true;}Base &operator=(const Base &a){if (this == &a) return *this;if (name != nullptr) delete[] name;int len = strlen(a.name);name = new char[len + 1];for (int i = 0; i < len; i++) name[i] = a.name[i];name[len] = '\0';num = a.num;return *this;}Base operator++(int){Base tmp(*this);num++;return tmp;}Base &operator++(){num++;return *this;}void print(){cout << "name = " << name << endl;cout << "num = " << num << endl;}private:int num;char *name;
};ostream&operator<<(ostream &os,const Base&p)
{os << "name = " << p.name << endl;os << "num = " << p.num << endl;return os;
}istream &operator>>(istream &in, Base &p)
{char c[10000];in >> c;int len = strlen(c);p.name = new char[len + 1];for (int i = 0; i < len; i++) p.name[i] = c[i];p.name[len] = '\0';in >> p.num;return in;
}Base operator+(const int a, Base& b)
{Base tmp(b);b.num += a;return tmp;
}int main()
{Base b("xiaom", 12);cout << b << endl;Base x("ADA", 32);x = x;Base c = b + x;cout << c << endl;cout<<x++<<endl;cout << ++x << endl;c = x;cout << c << endl;Base v = c + 32;int k = 3;Base x2 = c + k;Base x3 = k + c;cout << c << endl;cout << v << endl;v("hhh", 30);cout << v << endl;v = (c + 32);return 0;
}
