not quite the same issue, but you might be interested by
http://people.cs.uu.nl/stefan/blog/00012.html which notes:
Prelude> let apply = \f x -> f x
Prelude> let apply' = \f -> f
Prelude> apply undefined `seq` ()
()
Prelude> apply' undefined `seq` ()
*** Exception: Prelude.undefined
mulling this over helped me think about functions that were similar but
had different numbers of params. (the first only takes a function as its
first arg, the second, which I believe is the same as "id" takes
anything.)
t.
PR Stanley
a good sanity check for saneness of two fxs is to "quickcheck" them, as I
believe I provided an example to for a previous question of yours.
Though I think in this case that's impossible because, as someone else
pointed out, not even the function tyes agree.
t.
PR Stanley
Your intuition that id is related to const is a good one, since id can be defined in terms of const. Here is one of many: id' = flip const const [This is also called a CKK combinator. See http://en.wikipedia.org/wiki/S_combinator#Combinatory_calculi for more examples, such as SKK, SKS, and various other identity equivalents.] Whether these are "different" identity functions is an interesting question. Ask yourself how you could tell them apart. What if they both always gave the same answer, but one always took longer than the other? Are you allowed to "peek at the code" before deciding on equality? Does it even matter whether there is only one identity? [AKA, is function application faithful?] Interesting stuff. More at: http://en.wikipedia.org/wiki/Semiautomaton Dan PR Stanley wrote:
Hi is const = id? const 'x' 'y' 'x' id 'x' 'x'
Cheers, Paul
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participants (5)
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Bryan O'Sullivan -
Dan Weston -
Felipe Lessa -
PR Stanley -
Thomas Hartman