On Fri, Dec 30, 2011 at 9:20 AM, Colin Adams <colinpaulad...@gmail.com>wrote:
> > > On 30 December 2011 17:17, Gregg Reynolds <d...@mobileink.com> wrote: > >> >> On Dec 30, 2011, at 11:04 AM, Colin Adams wrote: >> >> >> >> On 30 December 2011 16:59, Gregg Reynolds <d...@mobileink.com> wrote: >> >>> >>> On Fri, Dec 30, 2011 at 12:49 AM, Heinrich Apfelmus < >>> apfel...@quantentunnel.de> 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. > Careful of circular reasoning here. Is f actually a "function" in the mathematical sense? It's that math sense that you need to reach your conclusion. BTW, the more I hear words like "clearly" and "obvious", the more I suspect that fuzziness is being swept under the carpet. - Conal
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