Channels have identity, so allocating a new one is a side effecting operation. Having it outside the IO monad would require (for example):
(newChan, newChan) = (let x = newChan in (x,x)) which is wrong. If you wrap newChan in unsafePerformIO then the compiler will feel free to apply rewrites like the above, which is unlikely to be what you wanted. Nick -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of S. Alexander Jacobson Sent: 23 April 2004 19:22 To: Haskell Mailing List Subject: [Haskell] Why is newChan in the IO Monad? Nothing actually happens when newChan is called except construction of a new datastructure. It would be nice to have non IO monad code be able to create a new Chan that gets passed to IO code that uses it somewhere else. Alternatively, is there a way to create a Chan outside the IO monad? -Alex- _________________________________________________________________ S. Alexander Jacobson mailto:[EMAIL PROTECTED] tel:917-770-6565 http://alexjacobson.com _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
