Re: [Haskell-cafe] On the purity of Haskell

On Fri, Dec 30, 2011 at 9:43 AM, Conal Elliott wrote:

On Fri, Dec 30, 2011 at 9:43 AM, Gregg Reynolds wrote:

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On Dec 30, 2011, at 11:20 AM, Colin Adams wrote: On 30 December 2011 17:17, Gregg Reynolds wrote:

On Dec 30, 2011, at 11:04 AM, Colin Adams wrote:

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On 30 December 2011 16:59, Gregg Reynolds wrote:

On Fri, Dec 30, 2011 at 12:49 AM, Heinrich Apfelmus <

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apfelmus@quantentunnel.de> wrote:

The function

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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?

Also: f is not a function, so I guess your proof is flawed.

Can you support the claim that f is not a function?