# Difference between revisions of "99 questions/Solutions/4"

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< 99 questions | Solutions

Danwizard208 (talk | contribs) (Added another two solutions) |
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myLength [] = 0 |
myLength [] = 0 |
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myLength (_:xs) = 1 + myLength xs |
myLength (_:xs) = 1 + myLength xs |
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+ | |||

+ | myLength' :: [a] -> Int |
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+ | myLength' list = myLength_acc list 0 -- same, with accumulator |
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+ | where |
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+ | myLength_acc [] n = n |
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+ | myLength_acc (_:xs) n = myLength_acc xs (n + 1) |
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</haskell> |
</haskell> |
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myLength' xs = snd $ last $ zip xs [1..] -- Just for fun |
myLength' xs = snd $ last $ zip xs [1..] -- Just for fun |
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myLength'' = snd . last . (flip zip [1..]) -- Because point-free is also fun |
myLength'' = snd . last . (flip zip [1..]) -- Because point-free is also fun |
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+ | myLength''' = fst . last . zip [1..] -- same, but easier |
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</haskell> |
</haskell> |
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## Revision as of 04:45, 24 June 2012

(*) Find the number of elements of a list.

```
myLength :: [a] -> Int
myLength [] = 0
myLength (_:xs) = 1 + myLength xs
myLength' :: [a] -> Int
myLength' list = myLength_acc list 0 -- same, with accumulator
where
myLength_acc [] n = n
myLength_acc (_:xs) n = myLength_acc xs (n + 1)
```

```
myLength' = foldl (\n _ -> n + 1) 0
myLength'' = foldr (\_ n -> n + 1) 0
myLength''' = foldr (\_ -> (+1)) 0
myLength'''' = foldr ((+) . (const 1)) 0
myLength''''' = foldr (const (+1)) 0
```

```
myLength' xs = snd $ last $ zip xs [1..] -- Just for fun
myLength'' = snd . last . (flip zip [1..]) -- Because point-free is also fun
myLength''' = fst . last . zip [1..] -- same, but easier
```

```
myLength = sum . map (\x -> 1)
```

This is `length`

in `Prelude`

.