必須返回對象時,別妄想返回其reference 【Effective C++ 條款21】
- 2020 年 5 月 20 日
- 筆記
class Rational { public: Rational(int numerator = 0, int denominator = 1) : n(numerator), d(denominator) { printf("Rational Constructor\n"); } ~Rational() { printf("Rational Destructor\n"); } Rational(const Rational& rhs) { this->d = rhs.d; this->n = rhs.n; printf("Rational Copy Constructor\n"); } private: int n, d; friend const Rational operator*(const Rational& lhs, const Rational& rhs); };
Rational的*運算符可以這樣重載:
const Rational operator*(const Rational& lhs, const Rational& rhs) { Rational tmp(lhs.n * rhs.n, lhs.d * rhs.d); return tmp; }
但是不可以這樣重載:【區別在於一個&】
const Rational& operator*(const Rational& lhs, const Rational& rhs) { Rational tmp(lhs.n * rhs.n, lhs.d * rhs.d); return tmp; }
當這樣去使用:
Rational x(1, 2), y(2, 3); Rational z = x * y;
第一種方法可以得到正確的結果,因為會調用Rational的拷貝構造函數將tmp賦給z,但是第二種方法返回的是tmp的引用,在函數退出前,tmp就被銷毀了,所以這樣做是不對的。
不過,第一種方法雖然功能上沒有問題,但是效率上有所欠缺,因為調用了三次構造函數,一次複製構造函數,一次析構函數
Rational Constructor
Rational Constructor
Rational Constructor
Rational Copy Constructor
Rational Destructor
可以進行返回值優化如下:
const Rational operator*(const Rational& lhs, const Rational& rhs) { return Rational(lhs.n * rhs.n, lhs.d * rhs.d); }
Rational Constructor
Rational Constructor
Rational Constructor
優化之後少調用了一次複製構造函數和析構函數
完整程式碼如下:
#include <stdio.h> class Rational { public: Rational(int numerator = 0, int denominator = 1) : n(numerator), d(denominator) { printf("Rational Constructor\n"); } ~Rational() { printf("Rational Destructor\n"); } Rational(const Rational& rhs) { this->d = rhs.d; this->n = rhs.n; printf("Rational Copy Constructor\n"); } private: int n, d; friend const Rational operator*(const Rational& lhs, const Rational& rhs); }; const Rational operator*(const Rational& lhs, const Rational& rhs) { return Rational(lhs.n * rhs.n, lhs.d * rhs.d); } int main() { Rational x(1, 2), y(2, 3); Rational z = x * y; return 0; }