从B打电话给A. [英] Call A from B
问题描述
鉴于A级和B级：
A级
{
公开：
void method_for_B_to_call（）;
....
};
如何让B回调method_for_B（） A而不让B知道A的定义（例如不使用#includeA.h），但只知道
A :: method_for_B_to_call（ ）回电？
谢谢！
Ele < El*@inter.com>在消息中写道
新闻：Ii ******************** @ bgtnsc05-news.ops.worldnet.att.net ... < blockquote class =post_quotes>鉴于A级和B级：
A级
{
公开：
void method_for_B_to_call（）;
.. 。
};
如何让B回调A的method_for_B（）而不让B知道A的定义（例如不使用#includeA .h"），但只知道A :: method_for_B_to_call（）回电？
谢谢！
我不喜欢除非method_for_B_to_call是静态的
方法，否则我认为这是可能的。是吗？
john
Ele写道：
鉴于A级和B级：
A级
公开：
void method_for_B_to_call（）;
...
};
如何让B回调A的method_for_B（）而不让B知道A的定义（例如不使用#includeA.h），但只知道
A :: method_for_B_to_call（）用于回拨？
谢谢！
您无法直接在定义中执行此操作。你需要为B类找到一个
抽象来使用;确切地说 - 取决于细节
您的问题。
这里有两种方法：
0）代表团：
//////标题（由B类使用）：
A类; //前方声明
//由B级使用：
C级{
A * pa_;
public：
void method_for_B_to_call（）;
};
//////.cpp文件：包括上面的标题和A类的定义
void C :: method_for_B_to_call（）{
return pa _-> method_for_B_to_call（）;
}
1）抽象基类（接口）A：
//由B使用，由A实现
class ABase {
public：
virtual~ABase（）= 0;
virtual void method_for_B_to_call（）= 0;
// ...
};
Denis
2004年5月26日星期三05 ：格林威治标准时间54:48，Ele < El*@inter.com>写道：
鉴于A级和B级：
A级
公开：
void method_for_B_to_call（） ;
...
};
如何让B回调A的method_for_B（）而不让B知道A的定义（例如不使用" #include" A.h"），但只知道A :: method_for_B_to_call（）用于回拨？
我似乎已经使用指向成员的工作了，尽管它是
凌晨3点而且我从来没有得过指针 - 成员工作
之前;-)
试试这个（我不认为宣布A是B'的好处的不完整类型
有资格作为让B知道A的定义。至少我希望它
不......）：
// bh：
A级;
B级
{
公开：
typedef void（A :: * ftype）（）;
void CallIt（A * ap，ftype f）;
};
// b.cpp：
#include" b.h"
void B :: CallIt（ A * ap，ftype f）
{
（ap-> * f）（）;
}
// amain.cpp：
#include< iostream>
使用std :: cout;
使用std :: endl;
#include" b.h"
A级
{
public：
void method_for_B（）{
std :: cout<< In method_for_B ...（） << std :: endl;
}
};
int main（）
{
void（A :: * fp）（）= A :: method_for_B;
A a1;
B b1;
b1.CallIt（& a1，fp）;
返回0;
}
输出：
方法_for_B ...（）
-leor
谢谢！
-
Leor Zolman --- BD软件--- www.bdsoft.com
C / C ++，Java，Perl和Unix的现场培训
C ++用户：下载BD Software的免费STL错误消息解密器：
www.bdsoft.com/tools/stlfilt.html
Given class A and class B:
class A
{
public:
void method_for_B_to_call();
....
};
How to make B to call back method_for_B() of A without letting B know the
definition of A (e.g. not using "#include "A.h"), but only know
A::method_for_B_to_call() for calling back?
Thanks!
"Ele" <El*@inter.com> wrote in message
news:Ii********************@bgtnsc05-news.ops.worldnet.att.net...Given class A and class B:
class A
{
public:
void method_for_B_to_call();
...
};
How to make B to call back method_for_B() of A without letting B know the
definition of A (e.g. not using "#include "A.h"), but only know
A::method_for_B_to_call() for calling back?
Thanks!
I don''t think that is possible unless method_for_B_to_call is a static
method. Is it?
john
Ele wrote:
Given class A and class B:
class A
{
public:
void method_for_B_to_call();
...
};
How to make B to call back method_for_B() of A without letting B know the
definition of A (e.g. not using "#include "A.h"), but only know
A::method_for_B_to_call() for calling back?
Thanks!
You cannot do it directly in your definitions. You need to find an
abstraction for class B to use; which exactly - depends on the details
of your problem.
Here are two approaches:
0) Delegation:
//////header (used by class B):
class A; //forward declaration
//used by class B:
class C {
A* pa_;
public:
void method_for_B_to_call();
};
//////.cpp file: includes the above header and the definition of class A
void C::method_for_B_to_call() {
return pa_->method_for_B_to_call();
}
1) Abstract base class (interface) for A:
//used by B, implemented by A
class ABase {
public:
virtual ~ABase() = 0;
virtual void method_for_B_to_call() = 0;
//...
};
Denis
On Wed, 26 May 2004 05:54:48 GMT, "Ele" <El*@inter.com> wrote:
Given class A and class B:
class A
{
public:
void method_for_B_to_call();
...
};
How to make B to call back method_for_B() of A without letting B know the
definition of A (e.g. not using "#include "A.h"), but only know
A::method_for_B_to_call() for calling back?
I seem to have gotten it to work using a pointer-to-member, despite it
being 3am and my never having ever gotten a pointer-to-member to work
before ;-)
Try this (I don''t think declaring A as an incomplete type for B''s benefit
would qualify as "letting B know the definition of A". At least I hope it
doesn''t...):
// b.h:
class A;
class B
{
public:
typedef void (A::*ftype)();
void CallIt(A *ap, ftype f);
};
// b.cpp:
#include "b.h"
void B::CallIt(A *ap, ftype f)
{
(ap->*f)();
}
// amain.cpp:
#include <iostream>
using std::cout;
using std::endl;
#include "b.h"
class A
{
public:
void method_for_B() {
std::cout << "In method_for_B...()" << std::endl;
}
};
int main()
{
void (A::*fp)() = A::method_for_B;
A a1;
B b1;
b1.CallIt(&a1, fp);
return 0;
}
Output:
In method_for_B...()
-leor
Thanks!
--
Leor Zolman --- BD Software --- www.bdsoft.com
On-Site Training in C/C++, Java, Perl and Unix
C++ users: download BD Software''s free STL Error Message Decryptor at:
www.bdsoft.com/tools/stlfilt.html
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