[Haskell-cafe] On the purity of Haskell

[Conal Elliott]

On Fri, Dec 30, 2011 at 9:20 AM, Colin Adams <colinpauladams at gmail.com>wrote:

>
>
> On 30 December 2011 17:17, Gregg Reynolds <dev at mobileink.com> wrote:
>
>>
>> On Dec 30, 2011, at 11:04 AM, Colin Adams wrote:
>>
>>
>>
>> On 30 December 2011 16:59, Gregg Reynolds <dev at mobileink.com> wrote:
>>
>>>
>>> On Fri, Dec 30, 2011 at 12:49 AM, Heinrich Apfelmus <
>>> apfelmus at 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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