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