[Haskell-cafe] loop problem

[Jochem Berndsen]

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 at functor.nl