一些赋值操作符重载的例子,我在网上看到的这个样子的:赋值运算符重载:void返回回访参考参数
#include <iostream>
using namespace std;
class Distance {
private:
int feet; // 0 to infinite
int inches; // 0 to 12
public:
// required constructors
Distance(){
feet = 0;
inches = 0;
}
Distance(int f, int i){
feet = f;
inches = i;
}
void operator = (const Distance &D) {
cout << "assigning..." << endl;
feet = D.feet;
inches = D.inches;
}
// method to display distance
void displayDistance() {
cout << "F: " << feet << " I:" << inches << endl;
}
};
int main() {
Distance D1(11, 10), D2(5, 11);
cout << "First Distance : ";
D1.displayDistance();
cout << "Second Distance :";
D2.displayDistance();
// use assignment operator
D1 = D2;
cout << "First Distance :";
D1.displayDistance();
return 0;
}
他们返回void从重载函数。如果D1是被调用的对象,这对我有意义。
其它实例返回类对象的引用。
#include <iostream>
using namespace std;
class Distance {
private:
int feet; // 0 to infinite
int inches; // 0 to 12
public:
// required constructors
Distance(){
feet = 0;
inches = 0;
}
Distance(int f, int i){
feet = f;
inches = i;
}
Distance& operator = (const Distance &D) {
cout << "assigning..." << endl;
feet = D.feet;
inches = D.inches;
return *this;
}
// method to display distance
void displayDistance() {
cout << "F: " << feet << " I:" << inches << endl;
}
};
int main() {
Distance D1(11, 10), D2(5, 11);
cout << "First Distance : ";
D1.displayDistance();
cout << "Second Distance :";
D2.displayDistance();
// use assignment operator
D1 = D2;
cout << "First Distance :";
D1.displayDistance();
return 0;
}
这没有意义,我(以第一个例子考虑时)。如果在第一个例子D1 = D2;
调用像D1.=(D2);
,为什么会在这种情况下,第二个例子中的工作?是不是像D1 = D1.=(D2);
?在一天结束时它有什么不同吗?
你是什么意思。 “+” ??? –
我的意思是+是函数名称。所以请拨打D1的+功能。哦,我明白了......它应该= ...... D1's =功能......我在编辑。 – JoeBass
http://en.cppreference.com/w/cpp/language/copy_assignment – zett42