Re: [boost] Review request: Proto

From: Eric Niebler (eric_at_[hidden])
Date: 2008-01-31 11:12:45

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