Hello Daniel, Thursday, December 11, 2008, 11:09:46 PM, you wrote: you is almost right. but ghc don't share results of function calls despite their type. it just assumes that value of any type may use a lot of memory even if this type is trivial :) example when automatic sharing is very bad idea is: main = print (sum[1..10^10] + sum[1..10^10])
Am Donnerstag, 11. Dezember 2008 16:18 schrieb Mattias Bengtsson:
The program below computes (f 27) almost instantly but if i replace the definition of (f n) below with (f n = f (n - 1) * f (n -1)) then it takes around 12s to terminate. I realize this is because the original version caches results and only has to calculate, for example, (f 25) once instead of (i guess) four times. There is probably a good reason why this isn't caught by the compiler. But I'm interested in why. Anyone care to explain?
main = print (f 27)
f 0 = 1 f n = let f' = f (n-1) in f' * f'
(compiled with ghc --make -O2)
Mattias
Not an expert, so I may be wrong. The way you wrote your function, you made it clear to the compiler that you want sharing, so it shares. With
g 0 = 1 g n = g (n-1)*g (n-1)
it doesn't, because the type of g is Num t => t -> t, and you might call it with whatever weird Num type, for which sharing might be a bad idea (okay, for this specific function I don't see how I would define a Num type where sharing would be bad). If you give g a signature like g :: Int ->> Int, the compiler knows that sharing is a good idea and does it (cool thing aside: with module Main where
f 0 = 1 f n = let a = f (n-1) in a*a
main = do print (f 27) print (g 30)
g 0 = 1 g n = g (n-1)*g (n-1)
main still runs instantaneously, but g n takes exponential time at the ghci prompt. That's because in main the argument of g is defaulted to Integer, so it's shared.) _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
-- Best regards, Bulat mailto:Bulat.Ziganshin@gmail.com