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

On Nov 28, 2006, at 5:49 PM, Peter Dimov wrote:

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.

The reasoning is (or was :) that using a placeholder in an expression does nothing more than delays the evaluation of that expression. So (&_1->*&aux::a)( x ) should have the same semantics as &x->*&aux::a This is how all other operators (+, -, ...) work in Lambda. In the case of ->* this seems in deed confusing, and potentially a less strict interpretation of this principle might be more useful. Cheers, Jaakko