It won't work because haskell functions can't have side-effects.
I'm not quite sure what you're trying to say here.
Jules
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 <[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.
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