Benoit wrote:
Hello,
I am having a look at your library which seems very interesting. But i am having trouble finding whether there would be a way to do the following using Proto :
typedef ... Variable; Variable x, y; evaluate( x - y );
The important thing that differs from your examples is that, in this case, x and y have the exact same type...
That doesn't matter. If Variable is a proto expression type, x-y will create a proto expression tree.
As far as i understand what you have done, this implies instantiating both an expr<> type and the associated context at the same time.
Sorry, I don't understand.
Such a process does not appear in any of your examples, and i can't see how it could be done the "Proto" way... Is it simply not possible ?
Given: template<typename Expr> typename proto::result_of::eval< Expr const, proto::default_context const
::type evaluate(Expr const &expr) { proto::default_context const ctx; return proto::eval(expr, ctx); }
You can do this: proto::literal<int> x(12), y(4); int z = evaluate(x - y); And it should display "8". Substitute your own context type in "evaluate()" and you can make it do anything you want. -- Eric Niebler Boost Consulting www.boost-consulting.com