条款21:必须返回对象时,别妄想返回其reference

问题代码：

//错误 inline const rational& operator*(const rational& lhs, const rational& rhs) { rational result(lhs.n * rhs.n, lhs.d * rhs.d); return result; } //极容易内存漏露 inline const rational& operator*(const rational& lhs, const rational& rhs) { rational *result = new rational(lhs.n * rhs.n, lhs.d * rhs.d); return *result; }

 

修改后代码：

inline const rational operator*(const rational& lhs, const rational& rhs) { return rational(lhs.n * rhs.n, lhs.d * rhs.d); }