On 9/15/10 10:39 PM, Conal Elliott wrote:
Hi Alex,
In Haskell, data structures cache, while functions do not.
Exactly. What this means is that when you call (slowFib 50) Haskell does not alter slowFib in any way to track that it maps 50 to $whatever; however, it does track that that particular expression instance evaluates to $whatever. This is why, when you define fib50=(slowFib 50) it only needs to compute $whatever the first time the value of fib50 is required. After the first evaluation it's the same as if we had defined fib50=$whatever. So, the function being memoized is the evaluation function of Haskell itself, not any of the functions _in_ the language. There are various libraries for memoizing functions in Haskell, they're just not part of the language definition. -- Live well, ~wren