[Haskell-cafe] problem with IO, strictness, and "let"

[Tim Chevalier]

On 7/13/07, Michael Vanier <mvanier at cs.caltech.edu> wrote:
> Stefan,
>
> Thanks for your comments, as always.
>
> What I meant by really-truly-absolutely-I-mean-right-now-seq is something that would evaluate its
> argument as far as it is possible to do so i.e. something that forces strict evaluation of an
> argument.  That's what I thought seq did, but now I see I was wrong; it only "goes one deep" as it
> were.  In fact, as you say, seq is not defined in terms of evaluation; all that it guarantees is
> that its first argument is either (a) bottom, in which the result of the entire seq is bottom, or
> (b) not bottom.  To do so it has to evaluate the first argument only far enough to show bottom-ness
> or not, which is not strict evaluation as I understand it.  So am I right in saying that Haskell has
> no way to force strict evaluation?  Or am I confused as to the correct definition of "strict"?
>

A function f is strict if f _|_ = _|_ -- seq, as defined in Haskell,
is strict in its first argument. So seq does force strict evaluation.
Even if you define "strict evaluation" as evaluating a structure
completely or returning _|_, then you can certainly force strict
evaluation in Haskell -- but you can't write a polymorphic function
that does so, unless you use type class overloading.

Cheers,
Tim

-- 
Tim Chevalier* catamorphism.org *Often in error, never in doubt
"'There are no atheists in foxholes' isn't an argument against
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