解引用运算符重载如何真正起作用？ [英] How does dereference operator overloading really work?

问题描述

对此有好感吗？这本教科书我已经为此花了很多钱？具体来说，我试图用2个级别取消引用

类型转换而不是1，即


X< T> - > z => Y'LT; T> - > z => T * - > z

和

* X< T> => * Y< T> => * T

Y< T>转换必须以表达式temp完成。在X的方法体内无法完成

。


我想我有点工作 - >运算符，但相同的

技术对*运算符不起作用。 *似乎消耗了

。

Joe Seigh

Is there a good write on this. The textbooks I have fluff over
on this? Specifically, I trying to dereference with 2 levels
of type conversion not 1, i.e.

X<T> -> z => Y<T> -> z => T* -> z
and
*X<T> => *Y<T> => *T
The Y<T> conversion has to be done as an expression temp. It cannot be done
inside a method body for X.

I think I sort of had it working for the -> operator but the same
technique doesn''t work for the * operator. The * appears to getting
consumed.

Joe Seigh

推荐答案

Joe Seigh写道：

Joe Seigh wrote:

这是否写得很好。这本教科书我对此嗤之以鼻？具体来说，我尝试用2级来取消引用类型转换而不是1，即，< T> X< X>。 - > z => Y'LT; T> - > z => T * - > z
和
* X< T> => * Y< T> => * T

Y< T>转换必须以表达式temp完成。在X的方法体内部无法完成。

我认为我有点工作 - >运算符，但相同的技术对*运算符不起作用。 *似乎被消耗了。

Is there a good write on this. The textbooks I have fluff over
on this? Specifically, I trying to dereference with 2 levels
of type conversion not 1, i.e.

X<T> -> z => Y<T> -> z => T* -> z
and
*X<T> => *Y<T> => *T
The Y<T> conversion has to be done as an expression temp. It cannot be done
inside a method body for X.

I think I sort of had it working for the -> operator but the same
technique doesn''t work for the * operator. The * appears to getting
consumed.

我真的没有得到你想要做的事情。也许张贴一大块o

代码主要编译并告诉我们你想做什么？

I don''t really get what you''re trying to do. Maybe posting a chunk o
code that mostly compiles and show us what you''d like to do ?

Joe Seigh写道：

Joe Seigh wrote:

这是否写得很好。这本教科书我对此嗤之以鼻？


呃......有人换了键盘上的''''和''''键吗？

具体来说，我试图取消引用具有2级的类型转换而不是1，即，X< T> - > z => Y'LT; T> - > z => T * - > z
和
* X< T> => * Y< T> => * T

Y< T>转换必须以表达式temp完成。在X的方法体内部无法完成。

我认为我有点工作 - >运算符，但相同的技术对*运算符不起作用。 *似乎消耗了
。

Is there a good write on this. The textbooks I have fluff over
on this?
Uh... Did somebody swap the ''?'' and ''.'' keys on your keyboard?
Specifically, I trying to dereference with 2 levels
of type conversion not 1, i.e.

X<T> -> z => Y<T> -> z => T* -> z
and
*X<T> => *Y<T> => *T
The Y<T> conversion has to be done as an expression temp. It cannot be done
inside a method body for X.

I think I sort of had it working for the -> operator but the same
technique doesn''t work for the * operator. The * appears to getting
consumed.

对不起，我完全不知道你在问什么。

-Kevin

-

我的电子邮件地址有效，但会定期更改。

要联系我，请使用地址来自最近的帖子。

I''m sorry, I have absolutely no idea what you are asking.

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.

Gianni Mariani写道：

Gianni Mariani wrote:

Joe Seigh wrote:

这是否写得很好。这本教科书我对此嗤之以鼻？具体来说，我尝试用2级来取消引用类型转换而不是1，即，< T> X< X>。 - > z => Y'LT; T> - > z => T * - > z
和
* X< T> => * Y< T> => * T

Y< T>转换必须以表达式temp完成。在X的方法体内部无法完成。

我认为我有点工作 - >运算符，但相同的技术对*运算符不起作用。 *似乎被消耗了。

Is there a good write on this. The textbooks I have fluff over
on this? Specifically, I trying to dereference with 2 levels
of type conversion not 1, i.e.

X<T> -> z => Y<T> -> z => T* -> z
and
*X<T> => *Y<T> => *T
The Y<T> conversion has to be done as an expression temp. It cannot be done
inside a method body for X.

I think I sort of had it working for the -> operator but the same
technique doesn''t work for the * operator. The * appears to getting
consumed.

我真的不知道你想要做什么。也许发布一大块代码，主要编译并告诉我们你想做什么？

I don''t really get what you''re trying to do. Maybe posting a chunk o
code that mostly compiles and show us what you''d like to do ?

T * X< T> :: operator - >（）;


提供一级转换。我不想要那个。像

Y< T> X< T> :: operator - >（）;

T * Y< T> :: operator - >（）;

我需要Y< T> ;温度生成。严格来说，我认为没有

重载 - >应该是需要的。类型转换应该足够

如果我有一个来自X< T>的转换路径了。到T *。


Joe Seigh

T * X<T>::operator ->();

gives one level of conversion. I don''t want that. Something like
Y<T> X<T>::operator ->();
T * Y<T>::operator ->();
I need a Y<T> temp generated. Strictly speaking, I would think no
overloading of -> should be needed. Type conversion should be sufficient
if I have a conversion path from X<T> to T*.

Joe Seigh

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