Bryan Burgers wrote:

On 7/30/07, peterv wrote:

...

Does Haskell support any form of automatic memorization?

For example, does the function

iterate f x

which expands to

[x, f(x), f(f(x)), f(f(f(x))), …

gets slower and slower each iteration, or can it take advantage of the fact that f is referentially transparent and hence can be "memoized / cached"?

Thanks, Peter

For 'iterate' the answer does not really need to be memoized.

Or, another way of phrasing that answer is 'yes'. The definition of iteration does memoize - although normally one would say 'share' - the intermediate results.

I imagine the definition of 'iterate' looks something like this:

iterate f x = x : iterate f (f x)

Haskell doesn't automatically memoize. But you are entitled to assume that named values are 'shared' rather than calculated twice. For example, in the above expression "x", being a named value, is shared between (a) the head of the list and (b) the parameter of the function "f" inside the recursive call to iterate. Of course sharing "x" may not seem very interesting, on the outermost call, but notice that on the next call the new "x" is the old "f x", and on the call after that the new "x" is "f (f x)" w.r.t the original "x". Jules

Hello peterv, Tuesday, July 31, 2007, 11:06:23 PM, you wrote: it is property of explicit *name* given to result of some expression. for example, when you write f x = g (x*x) (x*x) result of x*x isn't stored because it may be very large and compiler exactly follows your instruction - "calculate x*x two times" without trying to do optimization that may turn out to pessimization (of course, i mean that with *lazy* evaluation x*x is calculated only when needed and it may become a pessimization to save value between its usages as first and second argument) when you write f x = g t t where t=x*x compiler gets an instruction to calculate x*x only once and share calculated value between two parameters and it does just what you said

Thanks! Is this is also the case when using let and where, or is this just syntactic sugar?

-----Original Message----- From: Jules Bean [mailto:jules@jellybean.co.uk] Sent: Tuesday, July 31, 2007 5:09 PM To: Bryan Burgers Cc: peterv; haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] Newbie question about automatic memoization

Bryan Burgers wrote:

...

On 7/30/07, peterv wrote:

...

Does Haskell support any form of automatic memorization?

For example, does the function

iterate f x

which expands to

[x, f(x), f(f(x)), f(f(f(x))), .

gets slower and slower each iteration, or can it take advantage of the fact that f is referentially transparent and hence can be "memoized / cached"?

Thanks, Peter

For 'iterate' the answer does not really need to be memoized.

Or, another way of phrasing that answer is 'yes'. The definition of iteration does memoize - although normally one would say 'share' - the intermediate results.

...

I imagine the definition of 'iterate' looks something like this:

iterate f x = x : iterate f (f x)

Haskell doesn't automatically memoize. But you are entitled to assume that named values are 'shared' rather than calculated twice. For example, in the above expression "x", being a named value, is shared between (a) the head of the list and (b) the parameter of the function "f" inside the recursive call to iterate.

Of course sharing "x" may not seem very interesting, on the outermost call, but notice that on the next call the new "x" is the old "f x", and on the call after that the new "x" is "f (f x)" w.r.t the original "x".

Jules

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