Subject: Re: [Boost-users] [lamda] smart self binding of a function object
From: alfC (alfredo.correa_at_[hidden])
Date: 2010-05-11 13:45:05
> > The question is: What should be the type of XXXX and YYYY?
>
> You'll have to make operator() a template:
>
> template<class T>
> function<double(double)> operator()(const T& x_arg, double y_arg);
>
I was about to do that but then I though that doing so would open a
can of uncontrolled overload.
What if I want to restrict T to lambda expressions? what would be the
best way to do it, enable_if? some special tag? (Do lambda expression
have any special tag?)
Thank you,
Alfredo