boost bind to a data member callback behavior

Question

Can someone please explain this piece of code?

struct Class {
    boost::function<void()> member;
};
Class c;

boost::function<boost::function<void()>()> foo = boost::bind(&Class::member, &c);
boost::function<void()> bar = boost::bind(&Class::member, &c);

Why does the definition of bar compile and what is the result of it?

Edit: foo() works as expected, calling c.member(), but bar() doesn't.

Wow. That works? What compiler? I'd think that only `bar` works. — ltjax, Mar 30 '11 at 13:39
The second is easy, if you assume that the first is correct, since you can always just ignore the return type. — ltjax, Mar 30 '11 at 14:28

score 0 · Answer 1 · answered Mar 24 '11 at 13:00

You would need to bind if your class was like that:

class Class { public: void member(); };

Then what you want to do is that :

Class c;

boost::function<void()> the_function_i_want_to_call = boost::bind(&Class::member, c);

the_function_i_want_to_call.call();

score 0 · Accepted Answer · answered Mar 30 '11 at 14:38

The first call is used to "generate" an extractor functor. That functor, when called, will return the member that it was bound to.

The second call just hides the return type of the functor that is passed in (which is the same as in the first example). So essentially, calling bar will do nothing.

boost bind to a data member callback behavior

2 Answers2