On 30 December 2011 17:17, Gregg Reynolds <dev@mobileink.com> wrote:

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



On 30 December 2011 16:59, Gregg Reynolds <dev@mobileink.com> 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.