今天我注意到了一些东西。如果我创建三个版本的重载+运算符来处理每一个组合(对象+原始,原始+对象,对象+对象)一切都按预期执行:C++重载我的重载操作符?
class Int
{ int data;
public:
Int(){data = 0; };
Int (int size){ data = size; };
friend int operator+(Int, Int);
friend int operator+(int, Int);
friend int operator+(Int, int);
};
int operator+(Int lInt, Int rInt)
{ cout <<"first version. ";
return rInt.data + lInt.data;
}
int operator+(int lInt, Int rInt)
{ cout <<"second version. ";
return rInt.data + lInt;
}
int operator+(Int lInt, int rInt)
{ cout <<"third version. ";
return lInt.data + rInt;
}
int main(int argc, char *argv[]) {
Int int1 = 1;
cout << int1 + int1 <<endl; // prints "first version. 2"
cout << 3 + int1 << endl; // prints "second version. 4"
cout << int1 + 3 << endl; // prints "third version. 4"
}
但如果我删除第二和第三个版本中,它仍作品!?!
class Int
{ int data;
public:
Int(){data = 0; };
Int (int size){ data = size; };
friend int operator+(Int, Int);
};
int operator+(Int lInt, Int rInt)
{ cout <<"first version. ";
return rInt.data + lInt.data;
}
int main(int argc, char *argv[]) {
Int int1 = 1;
cout << int1 + int1 <<endl; // prints "first version. 2"
cout << 3 + int1 << endl; // prints "first version. 4"
cout << int1 + 3 << endl; // prints "first version. 4"
}
如何为我的重载+运营商,这意味着接受两个对象,能够采取一个int和对象。它如何能够采取对象和int?我希望我不会在这里忽略一些显而易见的东西!
也感谢你。现在我明白了! – 2015-04-04 20:14:22
不客气:) – bgoldst 2015-04-04 20:15:32