Re: [Haskell-cafe] how to break foldl' ?

Still a bit confused. My function is simply f x y = if x < 100 then x + y else x Sure the rest of y will not be touched(so if it is a file reading operation, no actual i/o will ever be performed) as they are not needed. But how would foldl knows that my logic won't need some other item in the list till the end ? So won't it call (f) X times even though all these X times just do a test and return ? --- Thomas Davie wrote:

No - lazy evaluation guarantees that if a reduct is never needed, it is never reduced. So as your function never needs the latter values in the list, it is never evaluated.

Bob

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