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Editorial Team
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Asked: 2026-05-26T00:45:15+00:00

Consider this: map fromEnum $ zipWith (==) aaaa abaa — [1,0,1,1] It would be

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Consider this:

map fromEnum $ zipWith (==) "aaaa" "abaa"
-- [1,0,1,1]

It would be nice to have only one step here:

zipWith (\x y -> fromEnum (x == y)) "aaaa" "abaa"

Now I can eliminate y:

zipWith (\x -> fromEnum.(x ==)) "aaaa" "abaa"

But I fail to eliminate x. Of course there are ways to “cheat”…

zipWith (curry (fromEnum . uncurry (==))) "aaaa" "abaa"

… but this looks uglier than the original lambda.

The function I look for would be somewhat similar to Data.Function.on, but “the other way around”. I have the feeling that there is an embarrassingly simple solution for this. Do I overlook something?

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1 Answer

  1. Editorial Team
    zipWith (\x -> fromEnum . (x ==)) "aaaa" "abaa"

    can be written as

    zipWith (\x -> (fromEnum .) (x ==)) "aaaa" "abaa"

    which can be written as

    zipWith ((fromEnum .) . (==)) "aaaa" "abaa"

    If you find this readable depends on taste I guess.

    EDIT: Another nice way to do it is with some combinators by Matt Hellige:

    zipWith ((==) $. id ~> id ~> fromEnum) "aaaa" "abaa"
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