Re: [Haskell-cafe] Do I have this right? Remembering Memoization!
I agree with what you meant, but not quite with what you said. To be pedantic: import Debug.Trace foo :: Int foo = trace Foo (bar 12) bar :: Int - Int bar x = trace Bar x main :: IO () main = foo `seq` foo `seq` return () main prints Foo\nBar\n showing that the bar is only evaluated once, because foo is already evaluated, even though it is referenced twice. So attempting to evaluate foo again just returns the same result. baz :: Int - Int baz x = trace Baz (bar x) correct :: IO () correct = baz 10 `seq` baz 11 `seq` return () Though, as you said, call, you probably meant foo was a function, and correct prints Baz\nBar\nBaz\nBar\n like you had indicated. But pedantically even the function: quux :: Int - Int quux x = trace Quux (bar 12) optmain :: IO () optmain = quux 10 `seq` quux 11 `seq` return () might print only once if GHC at the optimization level selected recognizes that quux doesn't depend on its argument and rewrote your code with more sharing. -Edward Kmett On Sun, Sep 13, 2009 at 7:45 PM, Mark Wotton mwot...@gmail.com wrote: On 14/09/2009, at 9:28 AM, Casey Hawthorne wrote: Do I have this right? Remembering Memoization! For some applications, a lot of state does not to be saved, since initialization functions can be called early, and these functions will remember - (memoize) their results when called again, because of lazy evaluation? You don't get memoisation for free. If you define a variable once in a where block, it's true that you'll evaluate it at most once, but if you repeatedly call a function foo that then calls bar 12 each time, bar 12 will be evaluated once per foo call. Cheers Mark ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Do I have this right? Remembering Memoization!
But pedantically even the function: quux :: Int - Int quux x = trace Quux (bar 12) optmain :: IO () optmain = quux 10 `seq` quux 11 `seq` return () might print only once if GHC at the optimization level selected recognizes that quux doesn't depend on its argument and rewrote your code with more sharing. Well to be specific, it depends on how you define function, quux :: Int - Int quux = trace Quux bar will print Quux once under the naive semantics. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Do I have this right? Remembering Memoization!
Do I have this right? Remembering Memoization! For some applications, a lot of state does not to be saved, since initialization functions can be called early, and these functions will remember - (memoize) their results when called again, because of lazy evaluation? -- Regards, Casey ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Do I have this right? Remembering Memoization!
On 14/09/2009, at 9:28 AM, Casey Hawthorne wrote: Do I have this right? Remembering Memoization! For some applications, a lot of state does not to be saved, since initialization functions can be called early, and these functions will remember - (memoize) their results when called again, because of lazy evaluation? You don't get memoisation for free. If you define a variable once in a where block, it's true that you'll evaluate it at most once, but if you repeatedly call a function foo that then calls bar 12 each time, bar 12 will be evaluated once per foo call. Cheers Mark ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
