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

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 < apfelmus@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.