On Sat, 29 Dec 2007 01:42:54 +0200, Jonathan Cast
<[EMAIL PROTECTED]> 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.
Well, that kind of loop won't work.
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