4

I want add a logger function to a worker class, how to pass a member function as a function pointer? use mem_fun?

here is the code sample:

class Work{
public:
    void (*logger) (const string &s);
    void do_sth(){if (logger) logger("on log")};
};

classs P{
public:
    void log(const string &s)(cout << s);
};

int main(){
    Work w;
    P p;
    w.logger = &p.log;
    w.do_sth();
}

edit:

I don't want to use void (P::*xxx)() because it stick to class P...

I know C++ hide sth, the real log function is: void log(P &p, const string &s),

and the real project is like this:

I create a CDialog, and there is a log function, it copy the log string to a CEdit.

So I need pass this log function to a Worker class, this class do some serial port job,

I need log and show the data send and recived...

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1
  • I think this is a very common design pattern. Commented Aug 4, 2010 at 3:02

3 Answers 3

Reset to default
5

You can accomplish this using std::function and std::bind:

#include <functional>
#include <iostream>
#include <string>

class Work {
public:
    std::function<void(const std::string&)> logger;
    void do_sth() { logger("on log"); }
};

class P {
public:
    void log(const std::string& s) { std::cout << s; }
};

int main() {
    Work w;
    P p;
    w.logger = std::bind(&P::log, p, std::placeholders::_1);
    w.do_sth();
}

Note that function and bind may not be in your implementation's standard library yet; you can also get them from the Boost libraries.

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7 Comments

@James McNellis, shouldn't std::placeholders::_1 be std::tr1::placeholders::_1 instead?
In my real project, I use(and compiled OK): using namespace std; device.logger = boost::bind(&CDeviceTesterDlg::logger, this, _1);
@Praetorian: It depends where you get bind from. Various older implementations have it and related names in namespace std::tr1, which was the place specified for them in the TR1 library extensions specification; the latest standard library releases have the names in namespace std, which is where they are specified to be located in (the forthcoming) C++0x standard.
@linjunhalida: For the sake of whomever might have to maintain your code, please, please, please don't use using namespace std;. There's no good reason to use it and it only causes trouble.
I take the suggestion, why don't use using namespace std? any discussion link?
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1

The only way to do it is as a static function. Unfortunately a static function isn't tied to a single object, but is global to the class. If you're lucky, the function taking a function pointer will also take a void*, which you can use to tie back to the original object - your static function can cast that back to an object pointer and call another function on the object to do the actual work.

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Comments

0

This would require closures, that C++ doesn't exactly have. You can make functors and use templates, but you'll have to go somewhat out of your way.

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1 Comment

@linjunhalida: James McNellis beat me to it. His answer is better.

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