I have this code:
filtro :: [(String, Int, Int)] -> [(String, Int, Int)]
filtro [] = []
filtro (x:xs)
| pior x xs = filtro xs
| otherwise = x : filtro xs
I need to make this variable static: pior x xs filtro xs . Any idea how?
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1What do you mean by "static"? If you mean it in either the C or the Java sense, then in either case the answer is "you don't, and there's a better way to do whatever you're thinking of".Silvio Mayolo– Silvio Mayolo2020-10-16 00:35:46 +00:00Commented Oct 16, 2020 at 0:35
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Static like keep the variable constant, without changing.John Smith– John Smith2020-10-16 00:37:13 +00:00Commented Oct 16, 2020 at 0:37
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1 Answer
It sounds like you want to apply pior to x and the whole list, not just the tail. I think some degree of separation of concerns could do you some favors. First off, your function is just filter, which can be defined as
filter :: (a -> Bool) -> [a] -> [a]
filter _ [] = []
filter p (x:xs)
| p x = x : filter x xs
| otherwise = filter x xs
So we can use that to our advantage. Then
filtro :: [(String, Int, Int)] -> [(String, Int, Int)]
filtro xs = filter (\x -> not (pior x xs)) xs
Since we make a closure around xs when it's the whole list, it'll remain the whole list in that closure for all of time (remember, in Haskell, variables don't change, ever).
If you really want to do it as one recursive function, you'll need an extra argument which remains the whole list for the duration of the computation.
filtro' :: [(String, Int, Int)] -> [(String, Int, Int)] -> [(String, Int, Int)]
filtro' _ [] = []
filtro' xss (x:xs)
| pior x xss = filtro xss xs
| otherwise = x : filtro xss xs
filtro :: [(String, Int, Int)] -> [(String, Int, Int)]
filtro xs = filtro' xs xs
In either case, you'll need to store the whole list somewhere, whether it's in an extra parameter or (preferred) transparently in a closure.
