Printing elements of a list on new lines
In most cases you don't need to program a loop over a list, it's been done already. To loop over a list with a monadic function, you would use mapM (and its mapM_ variant if you don't care about the result.)
If you use print, for ["1","2","3","4"]
you'd get :
Prelude> mapM_ print ["1","2","3","4"]
"1"
"2"
"3"
"4"
Prelude>
print is actually :
print :: Show a => a -> IO ()
print x = putStrLn (show x)
the show function causes the string "1"
to be converted to "\"1\""
, putStrLn prints that and a newline.
If you replace the print by putStrLn, you remove the conversion step and print directly the string:
Prelude> mapM_ putStrLn ["1","2","3","4"]
1
2
3
4
Prelude>
Now I would like to offer another solution. The Haskell way of doing things is doing as much as you can in a pure way, and only use IO when you need it.
So in this case we can join all strings to be printed with a \n
, and print all the strings at once.
To join all the strings there's a handy function : unlines
Prelude> unlines ["1","2","3","4"]
"1\n2\n3\n4\n"
Prelude>
Now you just have to print that; notice that unlines put a newline after the last item of the list, so we'll use putStr instead of putStrLn
Prelude> putStr ( unlines ["1","2","3","4"] )
1
2
3
4
Prelude>
Your function is:
printElements :: [String] -> IO()
printElements [] = return ()
printElements (x:xs) = do putStrLn x
printElements xs
If you already know about monads, you can use mapM_
function:
printElements :: [String] -> IO()
printElements = mapM_ putStrLn
Note: maybe you would have to read the chapter 8 of lyah.
Instead of explicit recursion, you can use mapM_
to call putStrLn
for every element. It works like the regular map
for lists, but is used with a monadic function (thus the "M"). The underscore variant is used when you only care about the side-effect (in this case, printing) and don't care about the result of the mapped function.
printElements :: [String] -> IO ()
printElements = mapM_ putStrLn