Talk:Y combinator: Difference between revisions
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(New page: ==Haskell stateless?== I don't know Haskell, but isn't the definition: y f = f (y f) '''not''' stateless as it seems to be defining y by referring to y. Haskell, no doubt has lazy evalua...) |
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'''not''' stateless as it seems to be defining y by referring to y. Haskell, no doubt has lazy evaluation to make it terminate, but the task does ask for a non-recursive, stateless definition of y. --[[User:Paddy3118|Paddy3118]] 09:18, 28 February 2009 (UTC) |
'''not''' stateless as it seems to be defining y by referring to y. Haskell, no doubt has lazy evaluation to make it terminate, but the task does ask for a non-recursive, stateless definition of y. --[[User:Paddy3118|Paddy3118]] 09:18, 28 February 2009 (UTC) |
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I googled [http://groups.google.co.uk/group/fa.haskell/browse_frm/thread/f0a62b6de1416d8b this]. --[[User:Paddy3118|Paddy3118]] 09:22, 28 February 2009 (UTC) |
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Revision as of 09:22, 28 February 2009
Haskell stateless?
I don't know Haskell, but isn't the definition:
y f = f (y f)
not stateless as it seems to be defining y by referring to y. Haskell, no doubt has lazy evaluation to make it terminate, but the task does ask for a non-recursive, stateless definition of y. --Paddy3118 09:18, 28 February 2009 (UTC)