Hello Szymon, Thursday, August 17, 2006, 12:18:25 PM, you wrote:
8.if (a == 0) && (b == 0) 9. then do 10. nr1 <- read (prompt "enter 1. number: ") 11. nr2 <- read (prompt "enter 2. number: ") 12. else do 13. let nr1 = a 14. nr2 = b {...}
1. as already said, your nr vars is local to the do blocks. you can't _assign_ to variables in Haskell, instead you should return _values_ that will become result of whole "if" expression: (nr1,nr2) <- if ... then do x <- .. y <- .. return (x,y) else do return (a,b) 2. as Chris said, "read" is a function (at least 'read' predefined in std Haskell library), while your 'prompt' should be I/O procedure. you can't call I/O procedures inside of functions, i.e. that is possible: function calls function I/O procedure calls function I/O procedure calls I/O procedure and that's impossible: function calls I/O procedure So you should assign result of procedure call to "variable" and then call function on this value: (nr1,nr2) <- if a==0 && b==0 then do x <- prompt "enter 1. number: " y <- prompt "enter 2. number: " return (read x, read y) else return (a,b) to be exact, x and y are not variables, but just bound identifiers like a and b. '<-' is special construct inside of 'do' block that binds to identifier value returned by I/O procedure call i've written tutorial on Haskell IO monad. you can try to read it, but it's more appropriate for intermediate Haskellers who has a good understanding of pure facilities of the language. but nevertheless try it - http://haskell.org/haskellwiki/IO_inside . i will be interesting to hear your opinion and depending on it will become more or less skeptical about suggesting it to Haskell newcomers fighting with mysterious IO monad :D in general, i suggest to learn pure foundations of Haskell such as lazy evaluation and higher-order functions. after that, learning IO monad using my tutorial will be as easy as saying "cheese" :) -- Best regards, Bulat mailto:Bulat.Ziganshin@gmail.com