
28 Dec
2007
28 Dec
'07
6:42 p.m.
On 28 Dec 2007, at 7:15 AM, Cristian Baboi wrote:
But I guess it won't work because the compiler will optimize it and the call will disappear.
On Fri, 28 Dec 2007 14:58:53 +0200, Cristian Baboi
wrote: Here is how I want print to be in Haskell
print :: (a->b) -> (a->b)
with print = id, but the following "side effect":
- I want to call the print function today, and get the value tomorrow.
Sorry, simply couldn't resist: Put a very long timing loop in the middle. jcc