Can pseq be defined in terms of seq?
As far as I know, seq ab evaluates (forces) a and b before returning b . It does not guarantee that a is evaluated first.
pseq ab evaluates a first, then evaluates/returns b .
Now consider the following:
xseq a b = (seq a id) b
Function application needs to evaluate the left operand first (to get a lambda form), and it can't blindly evaluate the right operand before entering the function because that would violate Haskell's non-strict semantics.
Therefore (seq a id) b must evaluate seq a id first, which forces a and id (in some unspecified order (but evaluating id does nothing)), then returns id b (which is b ); therefore xseq ab evaluates a before b .
Is xseq a valid implementation of pseq ? If not, what's wrong with the above argument (and is it possible to define pseq in terms of seq at all)?
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