Haskell Lazy Evaluation and Reuse

Posted by Jonathan Sternberg on Stack Overflow See other posts from Stack Overflow or by Jonathan Sternberg
Published on 2010-12-26T01:43:44Z Indexed on 2010/12/26 1:54 UTC
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Filed under:

haskell

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list-comprehension

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lazy-evaluation

I know that if I were to compute a list of squares in Haskell, I could do this:

squares = [ x ** 2 | x <- [1 ..] ]

Then when I call squares like this:

print $ take 4 squares

And it would print out [1.0, 4.0, 9.0, 16.0]. This gets evaluated as [ 1 ** 2, 2 ** 2, 3 ** 2, 4 ** 2 ]. Now since Haskell is functional and the result would be the same each time, if I were to call squares again somewhere else, would it re-evaluate the answers it's already computed? If I were to re-use squares after I had already called the previous line, would it re-calculate the first 4 values?

print $ take 5 squares

Would it evaluate [1.0, 4.0, 9.0, 16.0, 5 ** 2]?

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