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

On 11/05/2010 7:47 PM, alfC wrote:

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

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template<class T> function operator()(const T& x_arg, double y_arg);

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What if I want to restrict T to lambda expressions?

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

How can I do that? Can you point me to the right documentation if any.

In the context of Boost.Lambda there is no common base class for all expression templates (is there?), for example the placeholders do not inherit from anything. The only thing I found was the subclass 'sig' that seems to be defined for all expression types. But I am still scratching my head on how to expoit this 'sig' subclass.

Now that I think one obvious way to restrict T is to be an expression template that 'returns' a double (or convertible to 'double'), this information is contained in the sig subclass. But I am confused now to do any progress.

Alfredo

http://en.wikibooks.org/wiki/More_C%2B%2B_Idioms/Expression-template http://www.boost.org/libs/numeric/ublas/doc/expression_concept.htm C++ Expression Templates By Angelika Langer and Klaus Kreft, March 01, 2003 http://www.drdobbs.com/184401627 I have yet to tinker with expression templates in any meaningful way (aside from copy/pasting the example from wikibooks.org and playing with it), so beyond giving you those links as guides there's not much I can do. I do know that it's exactly what you want though and that the keys to your kingdom are the Var and Constant classes (or equivalent) from the wikibooks.org article. Good luck! Geoff