apfelmus wrote:
Bryan Donlan wrote:
evaluate x = (return $! x) >>= return
However, if >>= is strict on its first argument, then this definition is
no better than (return $! x).
According to the monad law
f >>= return = f
every (>>=) ought to be strict in its first argument, so it indeed seems
that the implementation given in the documentation is wrong.
But it is known that the monad laws only apply up to some weaker
equivalence than 'seq-equivalence'.
This has been discussed here countless times by people who understand it
better than me.
As I understand the summary the "=" sign in the monad laws mean
"represent identical actions, in terms of the effects produced and the
result returned". A kind of observational-equivalence for monad
execution, but weaker than direct equational equivalence in the presence
of seq.
(Some people view this as more of a bug in "seq" than in the monad laws)
Jules
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