成员函数指针困难 [英] Member function pointer woes
问题描述
我需要将指向成员函数的指针作为参数传递给函数
，它将指针指向函数作为一个论点。有什么方法可以这样做吗
除了重载函数吗？
这是我编写的一个小程序，用于检查是否可以获取<的地址br />
函数来自member-function-pointer，这样我就可以把它作为普通函数指针传递给
函数。
------------------------------------------------ ---
#include< stdio.h>
class classname {
public：
int add（int a，int b）{
printf（" inside add。\ ta =％d，b =％d \ n"，a，b）;
返回a + b;
}
} obj;
typedef int（classname :: * mfptr）（int， int）;
typedef int（* fptr）（int，int）;
main（）{
mfptr mem_func_ptr;
fptr func_ptr;
mem_func_ptr =& classname :: add;
func_ptr =（fptr）（& classname： ：add）;
printf（"& classname :: add is％llx \ n"，& classname :: add）;
printf（"& classname :: add is％llx \ n"，& classname :: add）;
printf（" mem_func_ptr是％llx \ n"，mem_func_ptr）;
printf（" func_ptr是％x \ n \ n"，func_ptr）;
printf（使用func_ptr添加调用。 sum是％d \ n \ n"，func_ptr（2,3））;
printf（" add with mem_func_fptr。sum is％d\\\
\ n"，
（obj。* mem_func_ptr）（2,3））;
}
--------- --------------------------------------------
我用-Wno-pmf-conversions选项编译它。这是节目的输出
：
---------------------- --------------------
& classname :: add is ffbeee78ff23e5d0
& classname :: add是ffbeee78ff243a4c
mem_func_ptr是ffbeee78ff243a4c
func_ptr是109bc
里面加。 a = 3，b = 68028
使用func_ptr调用add。总和是68031
里面加。 a = 2，b = 3
使用mem_func_fptr调用add。总和是5
----------------------------------- -------
几个问题：
1.为什么''& classname :: add''的值在两行不同？
2.当''classname :: add''用''func_ptr'调用时，为什么参数''b''的值会改变？
'？
3.如果这种方法不正确，那么请您告知
如何使用它？
谢谢。
Albert。
"阿尔伯特QUOT; <人**** @ iitg.ernet.in>在消息中写道
新闻：e9 ************************* @ posting.google.co m ... < blockquote class =post_quotes>
我需要将指向成员函数的指针作为参数传递给函数
，它将指针作为参数。除了重载函数之外，还有什么办法吗？
No.
[snip]
3.如果这种方法不正确，那么请你建议如何使用它？
写一个普通函数调用你的成员函数，然后传递
正常功能作为参数。
john
Albert写道：
我需要将指向成员函数的指针作为参数传递给
函数，该函数将指针作为参数。除了重载功能之外还有其它方法吗？
< snipped>
3.如果这种方法不正确，那么请您指教一下如何使用它？
我建议你考虑使用Boost.Function库。
http://www.boost.org/doc/html/function.html
我用它来创建一个库函数，它可以在不同的程序中使用指针到函数
或指向成员函数的指针。
Mark
" Albert" <人**** @ iitg.ernet.in> skrev i meddelandet
新闻：e9 ************************* @ posting.google.co m ... < blockquote class =post_quotes>我需要将指向成员函数的指针作为参数传递给函数
，它将指针作为参数。有什么方法可以这样做吗
除了重载函数？
你可以使用模板版的订阅者/发布者模式。
的实现并不漂亮，但使用方式有点整洁恕我直言;-)请注意
SubscriberBase和Subscriber可以重复用于其他的通知/>
相同的类型（对于任何类），在这种情况下，成员函数采用int
并返回void。
// Base订阅者类。必须是非模板，因为我们需要存储指向实例的指针。
class SubscriberBase
{
public：
virtual void exec（int）= 0;
};
//实际的订阅者类模板。存储指向对象的指针和
//指向其中一个成员函数。
template< class T> class Subscriber：public SubscriberBase
{
private：
typedef void（T :: * EventFunc）（int）;
T *目标; //指向收到通知对象的指针
EventFunc func; //指向要调用的成员函数的指针
public：
订阅者（T * target，EventFunc func）：func（func），target（target）{}
virtual void exec（int param）{（target-> * func）（param）; }
};
//这就是它的用法。
//做东西的类通知另一个对象有关某事......
类发布者
{
受保护：
SubscriberBase * sub; //可能真的是一个清单
public：
void registerSubscriber（SubscriberBase& s）
{
sub =& s;
};
void doSomething（int param）
{
// Do一些东西
// ...
sub-> exec（param）; // ...并通知
}
};
//测试类。收到通知。
class UserClass
{
public：
//启动整个事情
void start（）
{
//需要告诉UserClass这个实例的Publisher对象
某事
发布商测试;
//构建并注册订阅者对象。
订阅者< UserClass> sub（this，callback）;
test.registerSubscriber（sub）;
//调用发回通知的函数
test.doSomething（12345）;
}
//回调成员函数
void callback（int param）
{
cout<< 你好， << param<<结束;
}
};
Hi,
I need to pass a pointer-to-member-function as a parameter to a function
which takes pointer-to-function as an argument. Is there any way to do it
besides overloading the function?
Here is a small program I wrote to check if I could get the address of the
function from the member-function-pointer, so that I could pass it to the
function as a normal function-pointer.
---------------------------------------------------
#include<stdio.h>
class classname{
public:
int add(int a, int b){
printf("inside add. \ta = %d, b = %d\n", a,b);
return a+b;
}
}obj;
typedef int(classname::*mfptr)(int, int);
typedef int (*fptr)(int, int);
main(){
mfptr mem_func_ptr;
fptr func_ptr;
mem_func_ptr = &classname::add;
func_ptr = (fptr)(&classname::add);
printf("&classname::add is %llx\n", &classname::add);
printf("&classname::add is %llx\n", &classname::add);
printf("mem_func_ptr is %llx\n", mem_func_ptr);
printf("func_ptr is %x\n\n", func_ptr);
printf("add called with func_ptr. sum is %d\n\n", func_ptr(2,3));
printf("add called with mem_func_fptr. sum is %d\n\n",
(obj.*mem_func_ptr)(2,3));
}
-----------------------------------------------------
I compiled it with the -Wno-pmf-conversions option. Here is the output of
the program:
------------------------------------------
&classname::add is ffbeee78ff23e5d0
&classname::add is ffbeee78ff243a4c
mem_func_ptr is ffbeee78ff243a4c
func_ptr is 109bc
inside add. a = 3, b = 68028
add called with func_ptr. sum is 68031
inside add. a = 2, b = 3
add called with mem_func_fptr. sum is 5
------------------------------------------
A few questions:
1. Why is the value of ''&classname::add'' in the two lines different?
2. Why does the value of parameter ''b'' change when ''classname::add'' is
called with ''func_ptr''?
3. If this approach is not correct, then could you please advise as to how
to go about with it?
Thanks.
Albert.
"Albert" <al****@iitg.ernet.in> wrote in message
news:e9*************************@posting.google.co m...Hi,
I need to pass a pointer-to-member-function as a parameter to a function
which takes pointer-to-function as an argument. Is there any way to do it
besides overloading the function?
No.
[snip]
3. If this approach is not correct, then could you please advise as to how
to go about with it?
Write a normal function which calls your member function and then pass the
normal function as a parameter.
john
Albert wrote:Hi,
I need to pass a pointer-to-member-function as a parameter to a
function which takes pointer-to-function as an argument. Is there any
way to do it besides overloading the function?
<snipped>
3. If this approach is not correct, then could you please advise as
to how to go about with it?
I would advise you to consider using the Boost.Function library.
http://www.boost.org/doc/html/function.html
I have used it to have a library function which can take pointer-to-function
or pointer-to-member-function in different programs.
Mark
"Albert" <al****@iitg.ernet.in> skrev i meddelandet
news:e9*************************@posting.google.co m...I need to pass a pointer-to-member-function as a parameter to a function
which takes pointer-to-function as an argument. Is there any way to do it
besides overloading the function?
You could use a templated version of the subscriber/publisher pattern. The
implementation isn''t pretty, but the usage is kind of neat IMHO;-) Note that
SubscriberBase and Subscriber can be reused for other notifications of the
same type (for any class), in this case member functions that take an int
and return void.
// Base class for Subscriber. Must be a non-template since we
// need to store a pointer to an instance.
class SubscriberBase
{
public:
virtual void exec(int) = 0;
};
// The actual Subscriber class template. Stores a pointer to an object and a
pointer
// to one of it''s member functions.
template<class T> class Subscriber : public SubscriberBase
{
private:
typedef void (T::*EventFunc)(int);
T *target; // Pointer to the object to recieve notification
EventFunc func; // Pointer to member function to be called
public:
Subscriber(T *target, EventFunc func) : func(func), target(target) { }
virtual void exec(int param) { (target->*func)(param); }
};
// And this is how it''s used.
// Class that does stuff and notifies another object about something...
class Publisher
{
protected:
SubscriberBase *sub; // Could be a list really
public:
void registerSubscriber(SubscriberBase &s)
{
sub = &s;
};
void doSomething(int param)
{
// Do something
// ...
sub->exec(param); // ... and notify
}
};
// Test class. Recieves notifications.
class UserClass
{
public:
// Starts the the entire thing
void start()
{
// Publisher object that needs to tell this instance of UserClass about
something
Publisher test;
// Construct and register a subscriber object.
Subscriber<UserClass> sub(this, callback);
test.registerSubscriber(sub);
// Call a function that sends back notifications
test.doSomething(12345);
}
// Callback member function
void callback(int param)
{
cout << "Hello, " << param << endl;
}
};
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