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Editorial Team
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Asked: 2026-05-31T11:03:52+00:00

If I call the following Haskell code find_first_occurrence :: (Eq a) => a ->

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If I call the following Haskell code

find_first_occurrence :: (Eq a) => a -> [a] -> Int
find_first_occurrence elem list = (snd . head) [x | x <- zip list [0..], fst x == elem]

with the arguments

'X' "abcdXkjdkljklfjdlfksjdljjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj"

how much of the zipped list [('a',0), ('b',1), ] is going to be built?

UPDATE:

I tried to run

find_first_occurrence 10 [1..]

and returns 9 almost instantly, so I guess it does use lazy evaluation at least for simple cases? The answer is also computed “instantly” when I run

let f n = 100 - n
find_first_occurrence 10 (map f [1..])
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1 Answer

  1. Editorial Team

    All of it.

    Since StackOverflow won’t let me post such a short answer: you can’t get away with doing less work than looking through the whole list if the thing you’re looking for isn’t there.

    Edit: The question now asks something much more interesting. The short answer is that we will build the list:

    ('a',0):('b',1):('c',2):('d',3):('X',4):<thunk>

    (Actually, this answer is just the slightest bit subtle. Your type signature uses the monomorphic return type Int, which is strict in basically all operations, so all the numbers in the tuples above will be fully evaluated. There are certainly implementations of Num for which you would get something with more thunks, though.)

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