On Thu, 2009-01-15 at 13:40 +0100, Apfelmus, Heinrich wrote: > Eugene Kirpichov wrote: > > Well, your program is not equivalent to the C++ version, since it > > doesn't bail on incorrect input. > > Oops. That's because my assertion > > show . read = id > > is wrong. We only have > > read . show = id > show . read <= id (in the "less defined than" sense)
No, you only have read . show = id which often doesn't hold in practice. show . read </= id Assuming the first identity holds, you do of course have show . read . show = show and this probably holds even in most cases where read . show = id does not hold. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe