21 操作符重载
1 需要解决的问题
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复数运算
class Complex { public: int a; int b; }; int main() { Complex c1 = {1,2}; Complex c2 = {3,4}; Compelx c3 = c1 + c2; //error: no match for ‘operator+’ (operand types are ‘Complex’ and ‘Complex’) return 0; }
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示例:复数的加法操作
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Demo
#include <stdio.h> class Complex { int a; int b; public: Complex(int a = 0, int b = 0) { this->a = a; this->b = b; } int getA() { return a; } int getB() { return b; } friend Complex Add(const Complex& p1, const Complex& p2); }; Complex Add(const Complex& p1, const Complex& p2) { Complex ret; ret.a = p1.a + p2.a; ret.b = p1.b + p2.b; return ret; } int main() { Complex c1(1, 2); Complex c2(3, 4); Complex c3 = Add(c1, c2); // c1 + c2 printf("c3.a = %d, c3.b = %d\n", c3.getA(), c3.getB()); //c3.a = 4, c3.b = 6 return 0; }
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2 操作符重载
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C++ 中重载能够扩展操作符的功能
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操作符的重载以函数的方式进行
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本质:用特殊形式的函数扩展操作符的功能
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通过
operator
关键字可以定义特殊的函数 -
operator
的本质是通过函数重载操作符 -
语法
Type operator Sign(const Type& p1,const Type& p2) { Type ret; return ret; } Sign为系统预定义的操作符,如+,-,*,/等
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示例:操作符重载
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Demo
#include <stdio.h> class Complex { int a; int b; public: Complex(int a = 0, int b = 0) { this->a = a; this->b = b; } int getA() { return a; } int getB() { return b; } //全局操作符重载函数 friend Complex operator + (const Complex& p1, const Complex& p2); }; //全局操作符重载函数 Complex operator + (const Complex& p1, const Complex& p2) { Complex ret; ret.a = p1.a + p2.a; ret.b = p1.b + p2.b; return ret; } int main() { Complex c1(1, 2); Complex c2(3, 4); Complex c3 = c1 + c2; // 等价于函数调用形式:operator + (c1, c2) printf("c3.a = %d, c3.b = %d\n", c3.getA(), c3.getB()); //c3.a = 4, c3.b = 6 return 0; }
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3 成员函数重载操作符
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可以将操作符重载函数定义为类的成员函数
- 比全局操作符重载函数少一个参数(左操作数)
- 不需要依赖友元就可以完成操作符重载
- 编译器优先在成员函数中寻找操作符重载函数
class Type { public: Type operator Sign(const Tyep& p) { Type ret; return ret; } };
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示例:成员函数重载操作符
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Demo
#include <stdio.h> class Complex { int a; int b; public: Complex(int a = 0, int b = 0) { this->a = a; this->b = b; } int getA() { return a; } int getB() { return b; } //成员函数操作符重载 Complex operator + (const Complex& p) { Complex ret; printf("Complex operator + (const Complex& p)\n"); ret.a = this->a + p.a; ret.b = this->b + p.b; return ret; } friend Complex operator + (const Complex& p1, const Complex& p2); }; Complex operator + (const Complex& p1, const Complex& p2) { Complex ret; printf("Complex operator + (const Complex& p1, const Complex& p2)\n"); ret.a = p1.a + p2.a; ret.b = p1.b + p2.b; return ret; } int main() { Complex c1(1, 2); Complex c2(3, 4); Complex c3 = c1 + c2; //等价于函数调用形式: c1.operator + (c2) printf("c3.a = %d, c3.b = %d\n", c3.getA(), c3.getB()); return 0; }
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编译运行
Complex operator + (const Complex& p) c3.a = 4, c3.b = 6
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