Re: [Haskell-cafe] Why is this strict in its arguments?

On Tue, Dec 04, 2007 at 03:35:28PM -0800, Ryan Ingram wrote:

On 12/4/07, Stefan O'Rear wrote: Well, one usually says something like "f is strict in its 2nd argument" which on casual reading tends to make me think that it has something to do with the argument. By the actual definition, however, f _ _ = undefined is strict in all of its arguments; but it's clear from the definition that the arguments are irrelevant.

this becomes more clear if you translate strictness into a logical representation. f x y = z saying f is strict in y is equivalent to y diverges implies z diverges. note that this is a one way implication. so, if z diverges, then the implication is trivially satisfied, as it is if y doesn't diverge, however if y diverges then z must diverge and that is what strictness means. or equivalantly z diverges \/ ! y diverges for an example of why this is useful think of the following imagine we know we are applying a function f to _|_, and we know f is strict in its first argument. Knowing it is strict, we can elide the call to f altogether and return bottom immediately. this is true whether f examines it's argument or not, since f _|_ = _|_ and we know we are passing _|_ then we can just return _|_. This is one of many optimizations that 'strictness analysis' allows. John * note, I am using 'x diverges' and 'x is bottom' to mean the same thing. Not all agree this is correct usage, even sometimes I don't. but for the purposes of this it is fine IMHO. -- John Meacham - ⑆repetae.net⑆john⑈