How to use pointer to member function when when pointer to global function is required?

Question

I have the following problem. I have to use a function that takes a callback. Implementing the callback is the tricky part, because I need more information beyond that I can extract from the input parameters. I will try to give an example:

typedef int (*fptr) (char* in, char* out); // the callback i have to implement
int takeFptr(fptr f, char* someOtherParameters); // the method i have to use

The problem is that I need additional info except the "in" parameter to construct the "out" parameter. I tried this approach:

class Wrapper {
    public:
        int callback(char* in, char* out){
           // use the "additionalInfo" to construct "out"
        }
        char* additionalInfo;
}

...
Wrapper* obj = new Wrapper();
obj->additionalInfo = "whatIneed";
takeFptr(obj->callback, someMoreParams);

I get the following error from the compiler:

error: cannot convert 'Wrapper::callback' from type 'int (Wrapper::)(char*, char*)' to type 'fptr {aka int(*)(char*, char*)}'

Ugh. If there's no hope of changing the interface you need to use to take a std::function or something, then bind or similar can't help you. You'll just have to create a wrapper global function that knows where to find some instance of Wrapper and call callback on it. This sucks because it effectively forces you to have some kind of global data. — BoBTFish
– BoBTFish, Commented Aug 1, 2014 at 10:08
Even if you could convert from a pointer to member to a normal function pointer (you can't in portable code*), how do you expect the this pointer to be available to the callback? — Jonathan Wakely
– Jonathan Wakely, Commented Aug 1, 2014 at 10:11

user1095108 · Accepted Answer · 2014-08-01 10:17:49Z

4

You need to pass what you need to pass, in this case a pointer to a function.

::std::function<int (char*, char*)> forwardcall;

int mycallback(char* in, char* out) // extern "C"
{
  return forwardcall(in, out);
}

forwardcall can contain any functor, for example:

forwardcall = [obj](char* in, char* out){ return obj->callback(in, out); };

edited Aug 1, 2014 at 10:17

answered Aug 1, 2014 at 10:10

user1095108

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8 Comments

BoBTFish Over a year ago

Why extern "C"? There is no indication from the error it is needed.

user1095108 Over a year ago

@BoBTFish just in case it is needed.

Jonathan Wakely Over a year ago

if the fptr callback type isn't declared extern "C" then adding it to your callback "just in case" should be an error when you try to pass it to takeFPtr (although most compilers don't implement that rule)

user1095108 Over a year ago

@JonathanWakely I didn't know you could place extern "C" into a typedef. It is possible?

Jonathan Wakely Over a year ago

[dcl.link]/1 "Two function types with different language linkages are distinct types even if they are otherwise identical."

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