[Haskell-cafe] Re: evaluate vs seq

Michael Shulman viritrilbia at gmail.com
Wed Sep 13 16:24:42 EDT 2006

On 9/13/06, apfelmus at quantentunnel.de <apfelmus at quantentunnel.de> wrote:
> So `seq` forces its first argument. When we define
>   f x = x `seq` (Return x)
> we thereby get
>   f _|_    == _|_
>   f []     == Return []
>   f (x:xs) == Return (x:xs)
> To compare, the semantics of (evaluate) is
>   evaluate _|_    == ThrowException =/= _|_
>   evaluate []     == Return []
>   evaluate (x:xs) == Return (x:xs)
> That should answer your question.

I must not be phrasing my question very well; I feel like we're
talking past each other.  It seems to me that when writing actual
programs (rather than reasoning about denotational semantics) the
reason one would use `seq' or `evaluate' is to force something to be
evaluated "now" rather than "later", i.e. to get around Haskell's
default lazy execution.

Your semantics say that (x `seq` return x) and (evaluate x) have the
same result when x is anything other than _|_.  All well and good, but
(return x) *also* has those same results when x is not _|_.  Why would
one use the former two rather than (return x), if x is known not to be
_|_?  Because they evaluate x at different "times", right?  Even though
the eventual return value is the same, and thus the *semantics* are the
same.  So laying aside the formal semantics, what is the difference in
terms of actual, real, Haskell programs?


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