z****e
发帖数: 2024 | 1 class overb{
public:
virtual overb f(){}//invalide covariant type
};
class overd:public overb{
public:
virtual overd f(){}//invalide covariant type
};
再看：
class overb{
public:
virtual overb* f(){}//valide covariant type
};
class overd:public overb{
public:
virtual overd* f(){}//valide covariant type
};
难道overd 不是 overb吗？ 这不正好满足type 的 <=关系吗？
1. overd 对应 overb
2. overd* 对应 overb*
请问1, 和 2 ， 有什么不一样？为什么一个是Covariant Return Type,另一个不是呢
？ |
X****r
发帖数: 3557 | 2 Because an "overd *" can be used in place of an "overb *",
but an "overd" cannot be used directly in place of an "overb".
(check Liskov substitution principle)
Think about this:
overd d;
overb b = d;
And this:
overd* pd;
overb* pb = pd;
They are totally different.
【在 z****e 的大作中提到】
: class overb{
: public:
: virtual overb f(){}//invalide covariant type
: };
: class overd:public overb{
: public:
: virtual overd f(){}//invalide covariant type
: };
: 再看：
: class overb{
|
z****e
发帖数: 2024 | 3 五体投地。
【在 X****r 的大作中提到】
: Because an "overd *" can be used in place of an "overb *",
: but an "overd" cannot be used directly in place of an "overb".
: (check Liskov substitution principle)
: Think about this:
: overd d;
: overb b = d;
: And this:
: overd* pd;
: overb* pb = pd;
: They are totally different.
|
w****m
发帖数: 146 | 4 upcasting only applies to pointer
【在 z****e 的大作中提到】
: class overb{
: public:
: virtual overb f(){}//invalide covariant type
: };
: class overd:public overb{
: public:
: virtual overd f(){}//invalide covariant type
: };
: 再看：
: class overb{
|
z****e
发帖数: 2024 | 5 还有reference 啦。
【在 w****m 的大作中提到】
: upcasting only applies to pointer
|
