IO Bool -> Bool

[Wolfgang Jeltsch]

On Thursday, 2003-08-14, 07:20, CEST, Tn X-10n wrote:
> hai guys

> is it possible to convert IO Bool to Bool?

No. The reason for introducing the IO type is to preserve the purity of 
Haskell, i.e., to ensure that expression evaluation doesn't depend on the 
state of the outside world and doesn't alter this state.*) Allowing a 
conversion from IO t to t would nullify this.

As you might have noticed, there was (or still is) a discussion about "yet 
another monad tutorial" on The Haskell Cafe. Some of the messenges and, above 
all, the monad tutorial itself may help you understand how I/O in Haskell 
works. There is not a quick answer to your question like: "This way you 
convert an IO Bool to a Bool." You will have to do some reading to understand 
the basic ideas of I/O in Haskell. They are quiet different from what you 
might know from other programming languages.

Wolfgang

*) This is at least how I would formulate it, others would probably phrase it 
a bit different. ;-)

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