Difference between revisions of "Talk:Compose"
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(simplified standard solution) |
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Reading this page got me to write an arrow version of compose. Don't know if it should be included or not since it isn't a monadic solution.<p/> |
Reading this page got me to write an arrow version of compose. Don't know if it should be included or not since it isn't a monadic solution.<p/> |
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− | < |
+ | <haskell> |
composeArrow :: [a -> a] -> a -> a <br/> |
composeArrow :: [a -> a] -> a -> a <br/> |
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composeArrow = foldr ((>>>) . arr) (arr id) |
composeArrow = foldr ((>>>) . arr) (arr id) |
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− | </ |
+ | </haskell> |
<br>Or maybe this:<br> |
<br>Or maybe this:<br> |
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− | < |
+ | <haskell> |
composeArrow :: [a -> a] -> a -> a |
composeArrow :: [a -> a] -> a -> a |
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composeArrow = foldl ((>>>) . arr) returnA |
composeArrow = foldl ((>>>) . arr) returnA |
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− | </ |
+ | </haskell> |
+ | |||
+ | ---- |
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+ | |||
+ | The first version does not need the function composition, because the functions could be applied immediately. |
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+ | This leads to |
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+ | |||
+ | <haskell> |
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+ | compose = flip (foldl (flip id)) |
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+ | </haskell> |
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+ | Shall I replace the first solution? |
Revision as of 12:45, 22 October 2006
Reading this page got me to write an arrow version of compose. Don't know if it should be included or not since it isn't a monadic solution.
composeArrow :: [a -> a] -> a -> a <br/>
composeArrow = foldr ((>>>) . arr) (arr id)
Or maybe this:
composeArrow :: [a -> a] -> a -> a
composeArrow = foldl ((>>>) . arr) returnA
The first version does not need the function composition, because the functions could be applied immediately. This leads to
compose = flip (foldl (flip id))
Shall I replace the first solution?