Hi Roelof, On 05/10/2014 05:22 PM, Roelof Wobben wrote:
Is it valid Haskell if I change the putStrln to putStrln ( show n * n) so I do not have to use the let command. Another question if I change return n with return () does the loop ever end. My feeling says not because the value of n is never updated for the loop.
Careful! show n * n is equal to (show n) * n, which is not what you wanted. Function application binds very strongly in Haskell. putStrLn (show (n*n)) is what you intended. In Haskell, you cannot 'update' values. The loop will terminate, however. Let's see why -- loop 101 will evaluate to return (). This is the action that, when performed, will do nothing. loop 100 will evaluate to do { putStrLn (show 100); putChar ' '; putStrLn (show (100 * 100)); loop (100 + 1) } This is equal to do { putStrLn (show 100); putChar ' '; putStrLn (show (100 * 100)); return () } So `loop 100' is equal to the action that, when performed, will print 100 10000 loop 99 will evaluate to do { putStrLn (show 99); putChar ' '; putStrLn (show (99 * 99)); loop (99 + 1) } which is equal to the action that, when performed, will print 99 9801 100 10000 and so on. There is only one way to actually perform an action, which is to put it into the definition of 'main'. To summarize, in Haskell there is a strict separation between the *evaluation* of values (including actions such as the above) and the *execution* of I/O, whereas most other languages conflate the two. For more information, see http://www.haskell.org/haskellwiki/Introduction_to_IO As an aside, the way this code is written is fairly unidiomatic. Haskell programmers in general like to separate I/O if possible. We'd generally write: values n = [(x, x*x) | x <- [1..n]] and then a function that writes general pairs of values. HTH, Jochem -- Jochem Berndsen | jochem@functor.nl