Effective C++ Item 24 若所有参数皆需类型转换，请为此采用non-member函数

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经验：如果你需要为某个函数的所有参数（包括被this指针所指的那个隐喻函数）进行类型转换，那么这个函数必须是个non-member

示例：operator *是member函数的情况

#include <iostream>
#include <string>
using namespace std;

class Rational{
public:
Rational(int nn = 0, int dd = 1): n(n), d(dd){}
const Rational operator* (const Rational &rhs) const{return Rational(n * rhs.n, d * rhs.d);};
int numerator() const{return n;}
int denominator() const {return d;}
private:
int n, d;
};

int main(){
Rational oneHalf(1, 2);
Rational result;
result = oneHalf * 2; // ok
result = 2 * oneHalf; // error C2677: 二进制“*”: 没有找到接受“Rational”类型的全局运算符(或没有可接受的转换)
system("pause");
}

解析：

result= oneHalf * 2 等价于 result = oneHalf.operator*(2); 而int类型的2会被隐式类型转换为Rational类型

result= 2 * oneHalf 等价于 result = 2.operator*(oneHalf); 而int类型的并没有相应的class，也就没有operator*函数

纠正：让operator*成为non-member函数

#include <iostream>
#include <string>
using namespace std;

class Rational{
public:
Rational(int nn = 0, int dd = 1): n(n), d(dd){}
int numerator() const{return n;}
int denominator() const {return d;}
private:
int n, d;
};

const Rational operator* (const Rational &lhr, const Rational &rhs)
{
return Rational(lhr.numerator() * rhs.numerator(), lhr.numerator() * rhs.denominator());
};
int main(){
Rational oneHalf(1, 2);
Rational result;
result = oneHalf * 2;
result = 2 * oneHalf;
system("pause");
}