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 _______________________________________________ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
