C++運算子多載(5) 重載== explicit避免隱式轉換__C++

有的時候程式中存在隱藏式轉換

比如下面這個例子

#include<iostream>using namespace std;class myComplex {private:int real;   //複數的實部int image;  //複數的虛部public:myComplex(int real = 0, int image = 0) {this->real = real;this->image = image;}friend bool operator == (const myComplex &obj1, const myComplex &obj2);};bool operator == (const myComplex &obj1, const myComplex &obj2) {if (obj1.real == obj2.real && obj1.image == obj2.image)return true;return false;}int main() {myComplex TestA(3,0);myComplex TestB(10,0);if (TestA == 3) {cout << "SAME" << endl;}else {cout << "NOT SAME" << endl;}system("PAUSE");return 0;

我們通過重載操作符 == 來判斷兩個自訂的類:複數是否相等

myComplex(int real = 0, int image = 0)

在建構函式中,我們給形參定義了預設值 所以在產生對象時 如果不指定形參 便會使用預設值

C++中,如果建構函式可以只傳入一個參數來調用,則會發生隱式轉換

if (TestA == 3)

在上面行代碼中 將3傳入重載操作符函數 建構函式便會構造一個3,0的myComplex對象, 這便發生了隱式轉換(將int型轉為myComplex型)

我們可以通過使用explicit來避免隱式轉換

explicit myComplex(int real = 0, int image = 0) {this->real = real;this->image = image;}

這時必須指明類型

myComplex TestA(3,0);myComplex TestB(10,0);if (TestA == (myComplex)3) {    //必須指明類型cout << "SAME" << endl;}else {cout << "NOT SAME" << endl;}