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

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:

...

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

...

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