rampion

Washington, DC

del.icio.us/rampion

Age: 34

Mathematician, programmer, and researcher; with interests in algorithmic design, software engineering theory, and massively parallel computing. I'm also a keyboard junkie and an aspiring language nerd.

1d
awarded Good Question
Nov
23
awarded Notable Question
Nov
20
awarded Enlightened
Nov
20
awarded Nice Answer
Nov
20
revised How do I implement nested functions in haskell
added 158 characters in body
Nov
20
revised How do I implement nested functions in haskell
added 158 characters in body
Nov
20
revised How do I implement nested functions in haskell
added 202 characters in body
Nov
20
answered How do I implement nested functions in haskell
Nov
19
comment Is there a lazy Session IO Monad?
is7s: it's not obvious to me what that definition of (>>=) would be.
Nov
18
answered Is there a lazy Session IO Monad?
Nov
18
comment Make Data Type of Kind * -> * That's Not a Functor
and when you've figured that out, check out contravariant functors and bifunctors
Nov
15
accepted An Existing Size-Lazy Vector Type In Haskell
Nov
15
comment Marking duplicate lines
(it's probably because your colorscheme doesn't define the Repeat highlighting group)
Nov
15
comment Marking duplicate lines
Daps0l: try adding hi link Repeat Statement to your ~/.vimrc.
Nov
15
revised How to return a polymorphic type in Haskell based on the results of string parsing?
added 13 characters in body
Nov
15
comment How to return a polymorphic type in Haskell based on the results of string parsing?
It's probably also worth discussing how even though read :: Read a => String -> a dodges this issue by [1] inferring the returned type from context (not from its argument) and [2] hiding partiality through use of error :: String -> a (so written purely, it'd be more like readEither :: Read a => String -> Either String a from Text.Read).
Nov
15
answered How to return a polymorphic type in Haskell based on the results of string parsing?
Nov
14
comment Haskell delete Chars from String
Consider what would happen if f [] = [[]]. Then f "a" = f ('a':[]) = [] : map ('a':) (f []) = [] : map ('a':) ([[]]) = [] : ['a':[]] = [] : ["a"] = [ [], "a" ]
Nov
14
comment Haskell delete Chars from String
Jubobs: Yes, the empty list is not the same as the list containing the empty list. But they're both of the same type (otherwise you couldn't compare them for equality).
Nov
14
comment Haskell delete Chars from String
Jubobs: the empty list is also the empty list of lists :)
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