19 rivejä
731 B
TeX
19 rivejä
731 B
TeX
To cut the story short, the abstract solution looks like this:
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\begin{haskellcode}
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fold :: b -> (a -> b -> b) -> [a] -> b
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fold z f [] = z
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fold z f (x:xs) = f x (fold z f xs)
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\end{haskellcode}
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Whoa! What's going on here?\\
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Let's see...
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\begin{itemizep}
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\item \code{z} is what we return if the list is empty
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\item \code{f} is our function (e.g. \code{(*)} or \code{(+)})
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\item and the last remaining argument is the actual list we are working on
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\end{itemizep}
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\slidep
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The function application has the following form:\\
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\code{fold f z [a,b,c] == a `f` (b `f` (c `f` z))}
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\vspace{\baselineskip}
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\\
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This folds from the right, so the \emph{Prelude} already defines a function which is very similar to ours and called \textbf{foldr}. |