[Haskell-cafe] On the purity of Haskell

[Gregg Reynolds]

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
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