[phoenix][lambda] lambda to phoenix translation: get arguments from expression
Hi I have a phoenix expression of type actor< composite< divides_eval, vector< LambdaExp, composite< minus_eval, vector< argument<0>, value<double> > > > >
f_expr
with boost.lambda I could extract the different components (Lambda and
the actual value stored by value<double>) from f_expr as:
get<0>(f_expr.args) //for lambda
and
get<1>(get<1>(f_expr.args).args) //for lambda
what is the equivalent for Boost.phoenix?,
it seems that the phoenix actor doesn't have the member "args".
something like at_c<1>(at_c<1>(f_expr)) seems to be in the right
direction but it returns something of type value<double> from which I
don't know how to extract the value
in the same way at_c<1>(f_expr) returns the denominator, but it is not
an actor, so it is not a lambda expression.
the corresponding lambda type was:
lambda_functor
f_expr
Thank you, Alfredo
On 5/29/2010 11:29 AM, alfC wrote:
Check out the extension mechanism (somewhere near the end) where this is documented. See composites. Essentially, while lambda uses tuples, phoenix uses fusion containers -- more or less the same, but with more horsepower (i.e. algorithms and iterators). Regards, -- Joel de Guzman http://www.boostpro.com http://spirit.sf.net http://www.facebook.com/djowel Meet me at BoostCon http://www.boostcon.com/home http://www.facebook.com/boostcon
I read that part many times but I couldn't realize how to apply it to
this case.
I ended up calling
actor<LambdaExp>(at_c<0>(f_expr))
actor
On 5/29/2010 3:48 PM, alfC wrote:
Either works fine. Regards, -- Joel de Guzman http://www.boostpro.com http://spirit.sf.net http://www.facebook.com/djowel Meet me at BoostCon http://www.boostcon.com/home http://www.facebook.com/boostcon
participants (2)
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alfC -
Joel de Guzman