Re: [Boost-users] [lamda] smart self binding of a function object

On 11/05/2010 1:45 PM, alfC wrote:

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The question is: What should be the type of XXXX and YYYY?

You'll have to make operator() a template:

template<class T> function 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

What about forcing T to be or be derived from an expression template base class?