18 строки
854 B
TeX
18 строки
854 B
TeX
There is also a function that folds from the \emph{left} which is also in the \emph{Prelude} and called \textbf{foldl}.\\
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To summarize:
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\begin{haskellcode}
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foldr f z [a,b,c] == a `f` (b `f` (c `f` z))
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foldl f z [a,b,c] == ((z `f` a) `f` b) `f` c
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\end{haskellcode}
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For \hinline{foldl} the \hinline{z} is sort of the starting value.
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\vspace{\baselineskip}
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\\
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\pause
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We can even express foldl in terms of foldr and vice versa. If you are interested, have a look here:\\ \url{http://lambda.jstolarek.com/2012/07/expressing-foldl-in-terms-of-foldr/}
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\vspace{\baselineskip}
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\\
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You should definitely look them up in the Prelude and play with them: \url{https://hackage.haskell.org/package/base-4.8.0.0/docs/Prelude.html}
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\begin{haskellcode*}{bgcolor=mygrey,frame=single,numbers=none,label=GHCi}
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> foldr (-) 0 [1, 2, 3]
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> foldl (-) 0 [1, 2, 3]
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\end{haskellcode*} |