Creating a functor inside a member function without taking the class as an argument

Question

Creating a functor inside a member function without taking the class as an argument

Apologies for cryptographic decryption.

I want to create a functor of the following type:

const boost::function<bool ()>& functor

Please consider the class:

#include <boost/function.hpp>
class X { 
    public:
        bool foo();
        void bar() ;
};

void X::bar() {
    const boost::function<bool (X *)>& f = &X::foo;
}

bool X::foo() {
    std::cout << __func__ << " " << __LINE__ << " " << std::endl;
    return true;
}

I have:

const boost::function<bool (X *)>& f = &X::foo;

Is it possible something like

const boost::function<bool ()>& f = &X::foo;

with boost :: bind or something else?

thank

+3

c ++ boost functor bind

gudge June 15. 15 at 15:26

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1 answer

101010 · Accepted Answer · 2015-06-15T15:35:21+0000

A non-static member function must be called with an object. Thus, you should always implicitly pass a pointer this

as an argument.

You can accomplish this with boost::bind

:

const boost::function<bool()>& f = boost::bind(&X::foo, this);

Creating a functor inside a member function without taking the class as an argument

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