alexandre · March 10, 2015 02:36
diff --git a/haskell-beginners b/haskell-beginners
 Hi guys/girls
 <ubbersith> I'm trying to repeat a python test with Haskell...can someone help me? https://gist.github.com/alexandre/442ec9dc189e222d3aae
 <ubbersith> I didn't understand why the interpreter don't just evaluates the first acess (it is trying evaluate everything?)
 <bitemyapp> ubbersith: you created an infinite list.
 * firebird1 (~firebird@117.192.191.140) has joined #haskell-beginners
 <bitemyapp> ubbersith: actually, cyclic reference, sorry.
 <bitemyapp> > take 10 $ cycle [1, 2]
 <lambdabot>  [1,2,1,2,1,2,1,2,1,2]
 <ubbersith> ok, but my bool test (==) should works, shouldn't?
 <bitemyapp> no.
 <bitemyapp> the data structure the Python code creates isn't well-typed in Haskell, it has the infinite type
 <bitemyapp> so it doesn't even really make sense.
 <bitemyapp> even if you wrote code that constructed such a value, the type-checker would swat you.
 <ubbersith> hmmm, ok. Thanks =]
 <bitemyapp> ubbersith: there ways of expressing similar notions in Haskell without making something sensical.
 <bitemyapp> [(a, b, [c])] is well-typed.
 <bitemyapp> because really what you're expressing is a list of a tuple of a value, a value, and a list of values.
 <bitemyapp> but you lifted "more list" into the list in a way that doesn't make a lot of sense.
 <ubbersith> hmmm, I thought the interpreter was evaluate the last element, only when I access it.
 <bitemyapp> the problem is the definition itself
 <ubbersith> yeah,I know...that's just I test...something that I think curious...
 <ubbersith> (I know -> it not make sense)
 <bitemyapp> there are ways to define infinite lists, but a naively self-referencing definition like that is _|_
 <bitemyapp> neat code though, had me thinking for a moment with that Python code.
 <bitemyapp> I've used Python for a decade and never used that trick before.
 <ubbersith> =]
 <ski> ubbersith : there is no "identical object" comparision in Haskell
 <ubbersith> ski: hmmm, thank you man.
 * coltfred (~coltfred@71.15.210.194) has joined #haskell-beginners
 <ski> > let x = [1,2]; y = x ++ y in (drop 2 . drop 2 . drop 2 . drop 2) y == y  -- this hangs, as it tries to traverse the cyclic lists until it either hits the end, or a discrepancy
 <lambdabot>  mueval: ExitFailure 1
 <ski> ubbersith ^
 <ski> (btw, that `(drop 2 . drop 2 . drop 2 . drop 2) y' is just another way of writing `drop 2 (drop 2 (drop 2 (drop 2 y)))')
 <ski> (note that in my version, it's not the element at index `2' in `y' which is equal to `y', but rather it's the remainder of the list `y' after dropping the first two elements which is equal to `y' (though it can't be detected internally from Haskell), as suggested by the above evaluation being aborted due to timeout)
	Hi guys/girls
	<ubbersith> I'm trying to repeat a python test with Haskell...can someone help me? https://gist.github.com/alexandre/442ec9dc189e222d3aae
	<ubbersith> I didn't understand why the interpreter don't just evaluates the first acess (it is trying evaluate everything?)
	<bitemyapp> ubbersith: you created an infinite list.
	* firebird1 (~firebird@117.192.191.140) has joined #haskell-beginners
	<bitemyapp> ubbersith: actually, cyclic reference, sorry.
	<bitemyapp> > take 10 $ cycle [1, 2]
	<lambdabot> [1,2,1,2,1,2,1,2,1,2]
	<ubbersith> ok, but my bool test (==) should works, shouldn't?
	<bitemyapp> no.
	<bitemyapp> the data structure the Python code creates isn't well-typed in Haskell, it has the infinite type
	<bitemyapp> so it doesn't even really make sense.
	<bitemyapp> even if you wrote code that constructed such a value, the type-checker would swat you.
	<ubbersith> hmmm, ok. Thanks =]
	<bitemyapp> ubbersith: there ways of expressing similar notions in Haskell without making something sensical.
	<bitemyapp> [(a, b, [c])] is well-typed.
	<bitemyapp> because really what you're expressing is a list of a tuple of a value, a value, and a list of values.
	<bitemyapp> but you lifted "more list" into the list in a way that doesn't make a lot of sense.
	<ubbersith> hmmm, I thought the interpreter was evaluate the last element, only when I access it.
	<bitemyapp> the problem is the definition itself
	<ubbersith> yeah,I know...that's just I test...something that I think curious...
	<ubbersith> (I know -> it not make sense)
	<bitemyapp> there are ways to define infinite lists, but a naively self-referencing definition like that is _\|_
	<bitemyapp> neat code though, had me thinking for a moment with that Python code.
	<bitemyapp> I've used Python for a decade and never used that trick before.
	<ubbersith> =]
	<ski> ubbersith : there is no "identical object" comparision in Haskell
	<ubbersith> ski: hmmm, thank you man.
	* coltfred (~coltfred@71.15.210.194) has joined #haskell-beginners
	<ski> > let x = [1,2]; y = x ++ y in (drop 2 . drop 2 . drop 2 . drop 2) y == y -- this hangs, as it tries to traverse the cyclic lists until it either hits the end, or a discrepancy
	<lambdabot> mueval: ExitFailure 1
	<ski> ubbersith ^
	<ski> (btw, that `(drop 2 . drop 2 . drop 2 . drop 2) y' is just another way of writing `drop 2 (drop 2 (drop 2 (drop 2 y)))')
	<ski> (note that in my version, it's not the element at index `2' in `y' which is equal to `y', but rather it's the remainder of the list `y' after dropping the first two elements which is equal to `y' (though it can't be detected internally from Haskell), as suggested by the above evaluation being aborted due to timeout)
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