Re: [Boost-users] using lambda's pointer to memberfunctionandboost::function

Rodolfo Lima wrote:

...

function f = (&_1->*&aux::a)();

This doesn't work because you are evaluating the functor without arguments, whereas it requires one.

You are right, it doesn't work; lambda's operator->* seems a pretty odd beast. Interesting that nobody has discovered its quirks so far. FWIW, the syntax above is how it should work in order to be usable. Lambda currently implements aux x; (&_1->*&aux::a)( x )(); instead of (&_1->*&aux::a)()( x ); The current functionality doesn't make much sense to me; if you had x at the time the lambda expression is constructed, you could have used it directly and skip the _1 business.

...

function f = bind( &aux::a, _1 );

This would work, but I in my project I have to call operator->* because in my case aux is a class that models a pointer, but doesn't have a dereference operator, just operator->* and operator->, those are the only ways to access the containing pointer. Well, there's also a ptr member function that returns it, but I'm trying to achieve the most natural way to access the containing pointer members. By using aux's ptr member, I have to use bind functions, which is a little unnatural.

You can use the above syntax with boost::bind and overload get_pointer for aux, but this won't work for lambda::bind or std::tr1::bind. Can't you just add operator* to aux? A raw pointer can escape via ptr, so adding a way to dereference the pointer doesn't seem to make it any more dangerous.

...

or

function f = &aux::a;

That doesn't make it either because f(a) just would return a delayed call to &aux::a, not its actual result.

It works for your example and returns 5. When I add an aux_ptr class with a get_pointer overload,

...

function f = &aux::a;

works through it, too.