From: Matthew Vogt (mvogt_at_[hidden])
Date: 2004-02-18 02:07:43
Brian McNamara <lorgon <at> cc.gatech.edu> writes:
> Well, technically you don't even need lambda to implement monads. But
> to do monads well/easily, you need syntax sugar (along the lines of
> Haskell's "do" notation or FC++'s comprehensions). This sugar requires
> explicit lambda variables (the same kind you see in FC++ lambda's
> let/letrec). It's unclear to me if/how to do that with (or extend)
> boost::lambda in a natural way to get this.
So, the less-concise syntax of fcpp::lambda means that there's more often
a term in an expression which can be used to cause an operator overload to
be selected from the lambda library?
> This is a pretty rambling and vague-sounding answer, I suppose, hrm.
Depends on whether you agree with my attempt to re-express it, I guess :)
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