[Haskell-cafe] On the purity of Haskell

[Conal Elliott]

On Fri, Dec 30, 2011 at 9:43 AM, Gregg Reynolds <dev at mobileink.com> wrote:

> On Dec 30, 2011, at 11:20 AM, Colin Adams 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.
>>>>
>>>>  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.
>
>
> That's called begging the question.  f is not a function, so I guess your
> proof is flawed.
>
> It seems pretty clear that we're working with different ideas of what
> constitutes a function.  When I use the term, I intend what I take to be
> the standard notion of a function in computation: not just a unique mapping
> from one input to one output, but one where the output is computable from
> the input.  Any "function" that depends on a non-computable component is by
> that definition not a true function.  For clarity let's call such critters
>  quasi-functions, so we can retain the notion of application.  Equality
> cannot be defined for quasi-functions, for obvious reasons.
>
> f is a quasi-function because it depends on getAnIntFromUser, which is not
> definable and is obviously not a function.  When applied to an argument
> like 42, it yields another quasi-function, and therefore "f 42 = f 42" is
> false, or at least unknown, and the same goes for f 42 != f 42 I suppose.
>
> -Gregg
>

Please don't redefine "function" to mean "computable function". Besides
distancing yourself from math, I don't think doing so really helps your
case.

And on what do you base your claim that getAnIntFromUser is not definable?
Or that applying it (what?) to 42 gives a quasi-function?
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://www.haskell.org/pipermail/haskell-cafe/attachments/20111230/db643cae/attachment.htm>