At 11:15 AM -0500 2003/10/31, Harris, Andrew wrote:
> Hi -
> I am trying to work out how the following function using "fix" is
> evaluated. I am hoping someone could look at my step-by-step breakdown
> of how I think evaluation works and see if I'm correct. My main
> question is how the order of operation (fixity?) is understood in going
> from step [3] to [4], if this is indeed how the evaluation would take
> place.
>
> Any help is appreciated,
> -andrew
>
> Here's the Haskell function:
>
> ---
> fix f = f (fix f)
>
> wierdFunc2 x y z = if y - z > z then x (y-z) z
> else y - z
>
> myRemainder = fix wierdFunc2
> ---
>
> Here's my "evaluation":
You're pretty close. I've noted a couple of differences below.
> myRemainder 12 5 -- [1]
> =
> fix wierdFunc2 12 5 -- [2] by substitution
> =
> wierdFunc2 fix wierdFunc2 12 5 -- [3] apply "fix"
wierdfunc2 (fix wierdFunc2) 12 5
> =
> wierdFunc2 myRemainder 12 5 -- [4] by substitution (?)
No, equations "rewrite" only "from left to right". Skip the step above
and the step below.
> =
> myRemainder 7 5 -- [5] apply "wierdFunc2"
> =
> fix wierdFunc2 7 5 -- [6] by substitution
> =
> wierdFunc2 fix wierdFunc2 7 5 -- [7] apply "fix"
Similarly, the above needs (fix wierdFunc2) to be parenthesized.
> =
> wierdFunc2 myRemainder 7 5 -- [8] by substitution (?)
Skip the above step.
> =
> 2 -- [9] apply "wierdFunc2"
Dean
