When I tried the following code in cghi:
take 1 $ take 1 $ repeat [1..]
I was expecting the result of 1 instead of [[1,2,3,4,5,6,7,8,9,10,... printing on my terminal.
Why is lazy evaluation not functioning as I'm hoping under such situation?
Why is lazy evaluation in Haskell "not being lazy"?
119 views Asked by Qi Ge At
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There are 2 answers
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takeis of typeInt -> [a] -> [a], i.e. it returns a list. It seems you’re looking forhead, which returns one element.