Re: [Haskell-cafe] Strange behavior with listArray

On Montag, 12. November 2012, 08:36:49, Bas van Dijk wrote:

On 12 November 2012 04:50, Alex Stangl wrote:

...

I'm stymied trying to figure out why the program below blows up with <<<loop>>> when I use "f 0"

If you replace the a!0 in f by its value 0, f is equivalent to:

f k = if k > 0 then f 0 else 0 : f 1

Do you see the loop now?

I see no loop in that, and ghci doesn't either: Prelude> let f :: Int -> [Int]; f k = if k > 0 then f 0 else 0 : f 1 Prelude> take 5 $ f 1 [0,0,0,0,0] and if you use (f 0) instead of (f (a!0)) there, it works.

Maybe you meant f to be:

f k = if k > 0 then f (a!k) else 0 : f 1

Loops too. The problem, Alex, is that f k = if k > 0 then f (a!0) else 0 : f 1 is strict, it needs to know the value of (a!0) to decide which branch to take. But the construction of the array a needs to know how long the list (0 : f 0) is (well, if it's at least four elements long) before it can return the array. So there the cat eats its own tail, f needs to know (a part of) a before it can proceed, but a needs to know more of f to return than it does. g and h are not strict, they simply let the construction write thunks into the array elements, and those can then later be evaluated after the construction of a has returned.