在C++进行运算符重载时, 一般来讲,运算符两边的对象的顺序是不能交换的. 比如下面的例子:
#include <iostream> using namespace std; class Distance { private: int feet; // 0 到无穷 int inches; // 0 到 12 public: // 所需的构造函数 Distance(){ feet = 0; inches = 0; } Distance(int f, int i){ feet = f; inches = i; } // 显示距离的方法 void displayDistance() { cout << "F: " << feet << " I:" << inches <<endl; } // 重载加运算符( + ) Distance operator+ (int i) { feet = feet + i; inches = inches + i; return Distance(feet, inches); } }; int main() { Distance D1(11, 10), D2(5, 11); Distance D3 , D4; D3 = 3 + D1; D4 = D1 + 3; D3.displayDistance(); D4.displayDistance(); if( D2 < D1 ) { cout << "D1 is less than D2 " << endl; } else { cout << "D2 is less than D1 " << endl; } return 0; }
在重载加号'+'运算符时, D4 = D1 + 3 是正确的, 而D3 = 3 + D1 提示出错.
解决该问题的一个方法是声明一个友元函数. 如下所示:
#include <iostream> using namespace std; class Distance { private: int feet; // 0 到无穷 int inches; // 0 到 12 public: // 所需的构造函数 Distance(){ feet = 0; inches = 0; } Distance(int f, int i){ feet = f; inches = i; } // 显示距离的方法 void displayDistance() { cout << "F: " << feet << " I:" << inches <<endl; } // 重载负运算符( - ) Distance operator+ (int i) { feet = feet + i; inches = inches + i; return Distance(feet, inches); } friend Distance operator+ (int i, const Distance& D) { Distance d; d.feet = D.feet + i; d.inches = D.inches + i; return Distance(d.feet, d.inches); } }; int main() { Distance D1(11, 10), D2(5, 11); Distance D3 , D4; D3 = 3 + D1; D4 = D1 + 3; D3.displayDistance(); D4.displayDistance(); if( D2 < D1 ) { cout << "D1 is less than D2 " << endl; } else { cout << "D2 is less than D1 " << endl; } return 0; }
这时便能过通过编译,顺利执行了.