On Aug 17, 2006, at 10:18 AM, Szymon Ząbkiewicz wrote:

Hi.

When trying to compilke this code:

{...} 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 {...}

The compiler tells me thats there's an error on line 10: "The last statement in a 'do' construct mesy be an expression"

Could you tell me how to change it so that the "declaration" of the first nr1 and nr2 is still in the "then" block.

The problem is that variables defined in the branch of an if are local to the branch. If you want to use them outside you have to return them from the branch: do (nr1, nr2) <- if (a == 0) && (b == 0) then do nr1 <- read (prompt "enter 1. number: ") nr2 <- read (prompt "enter 2. number: ") return (nr1, nr2) else do let nr1 = a nr2 = b return (nr1, nr2) / Ulf

Szymon Ząbkiewicz wrote:

Hi.

When trying to compilke this code:

{...} 8.if (a == 0) && (b == 0) 9. then do 10. nr1 <- read (prompt "enter 1. number: ") 11. nr2 <- read (prompt "enter 2. number: ")

The nr2 here is not passed to the rest of the do block started on line 9

12. else do 13. let nr1 = a 14. nr2 = b

The nr1 and nr2 in the else block have absolutely nothing to do with the nr1 and nr2 from the then block. The names are the same, but that does not make them the same as they could be totally different types.

{...}

The compiler tells me thats there's an error on line 10: "The last statement in a 'do' construct mesy be an expression"

Could you tell me how to change it so that the "declaration" of the first nr1 and nr2 is still in the "then" block.

The "x <- foo" syntax is not a declaration. Also, the type of "read" is String->a which is NOT a monad type, so I will fix that as well: One could do this: (nr1,nr2) <- if (a==0) && (b==0) then do a' <- liftM read (prompt "...") b' <- liftM read (prompt "...") return (a',b') else return (a,b) The ghc compiler is usually smart enough to remove the tuple (,) construction from the code.

Szymon Z??bkiewicz wrote:

The compiler tells me thats there's an error on line 10: "The last statement in a 'do' construct must be an expression"

I think, you have reached the point where treating do-notation as magic won't help you. Remember,

do nr1 <- read (prompt "enter 1. number: ") nr2 <- read (prompt "enter 2. number: ")

is syntactic sugar for

read (prompt "enter 1. number: ") >>= \nr1 -> read (prompt "enter 2. number: ") >>= \nr2 ->

and it obvious that something is missing after the last arrow. That's the expression the compiler is complaining about. After the translation, it is also completely clear, that there is no "variable" which is ever "declared" and could be "assigned". On a side note, using "trap values" like the special 0 is an ugly style inherited from C. You might want to get used to explicit representations for missing values. Compare this:

read_new :: Maybe (Int, Int) -> IO (Int, Int) read_new (Just ab) = return ab read_new Nothing = do n1 <- read_prompt "enter 1. number: " n2 <- read_prompt "enter 2. number: " return (n1, n2) where read_prompt p = prompt p >>= readIO

Also note the 'read_prompt' function; I'm pretty sure you got the types of 'prompt' and 'read' messed up, too. So in anticipation of your next question: 'read'ing the 'prompt' action is not the same as 'read'ing the result of the 'prompt' action. Only the latter makes sense. Udo. -- "Enthusiasm is contagious, and so is boredom." -- Paul Graham