rpglover64 at gmail.com
Wed Sep 15 14:03:26 EDT 2010
I feel that there is something that I don't understand completely: I have
been told that Haskell does not memoize function call, e.g.
> slowFib 50
will run just as slowly each time it is called. However, I have read that
Haskell has call-by-need semantics, which were described as "lazy evaluation
I understand that
> fib50 = slowFib 50
will take a while to run the first time but be instant each subsequent call;
does this count as memoization?
(I'm trying to understand "Purely Functional Data Structures", hence this
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