On 30 December 2011 17:17, Gregg Reynolds <[email protected]> wrote:
> > On Dec 30, 2011, at 11:04 AM, Colin Adams wrote: > > > > On 30 December 2011 16:59, Gregg Reynolds <[email protected]> wrote: > >> >> On Fri, Dec 30, 2011 at 12:49 AM, Heinrich Apfelmus < >> [email protected]> wrote: >> >>> >>> The function >>> >>> f :: Int -> IO Int >>> f x = getAnIntFromTheUser >>= \i -> return (i+x) >>> >>> is pure according to the common definition of "pure" in the context of >>> purely functional programming. That's because >>> >>> f 42 = f (43-1) = etc. >>> >>> Put differently, the function always returns the same IO action, i.e. >>> the same value (of type IO Int) when given the same parameter. >>> >> >> >> >> time t: f 42 (computational process implementing func application >> begins…) >> t+1: <keystroke> = 1 >> t+2: 43 (… and ends) >> >> time t+3: f 42 >> t+4: <keystroke> = 2 >> t+5: 44 >> >> Conclusion: f 42 != f 42 >> >> (This seems so extraordinarily obvious that maybe Heinrich has something >> else in mind.) >> >> This seems such an obviously incorrect conclusion. > > f42 is a funtion for returning a program for returning an int, not a > function for returning an int. > > > My conclusion holds: f 42 != f 42. Obviously, so I won't burden you with > an explanation. ;) > > -Gregg > Your conclusion is clearly erroneous. proof: f is a function, and it is taking the same argument each time. Therefore the result is the same each time.
_______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
