Converting user input (Integer) to conditional print statements

Question

I am trying to get a numeric value from the user (ranging 1-10) and using conditional statements (if number >= 1 && <=3) print out health status (e.g. putStrLn "your health is poor") but getting error message that I can't go past

health :: IO ()
health = do 
        putStrLn "State your health using numbers 1 - 10: "
        num <- getLine
        --putStrLn "Your health is: "
        if num >=1 && <=3
              then 
               do putStrLn "Your health is poor"
           else if num >=4 && getLine <=7
              then putStrLn "Your health is OK"
           else if num >=8 && getLine<=10
              then putStrLn "your health is fanstastic"
              else "Wrong health range indicated"

Error message:

healthCheck.hs:9:1: warning: [-Wtabs]
Tab character found here, and in five further locations.
Please use spaces instead.
|
9 |                         --putStrLn "Your health is: "   | ^^^^^^^^

healthCheck.hs:10:39: error: parse error on input ‘<=’
|
10 |                         if num >=1 && <=3    |

Alexey Raga · Accepted Answer · 2019-05-17 12:27:35Z

7

One immediate problem is that you write:

if num >=1 && <=3

instead of

if num >=1 && num <= 3

But even then it will not work since the type of num is String (since you use getLine :: IO String, and you cannot compare it with numeric values.

You would need to use something like read to "read" the string into a numeric value (note that it'll crash if the string does not represent a numeric value).

Next, expressions like getLine <= 7 will not work either, for the same reason: getLine :: IO String, so you cannot compare IO String with a numeric value.

This code will work:

health :: IO ()
health = do
            putStrLn "State your health using numbers 1 - 10: "
            str <- getLine
            let num = read str
            --putStrLn "Your health is: "
            if (num >=1 && num <=3 )
               then putStrLn "Your health is poor"
               else if num >=4 && num <=7
                  then putStrLn "Your health is OK"
               else if num >=8 && num <=10
                  then putStrLn "your health is fanstastic"
                  else putStrLn "Wrong health range indicated"

However I'd recommend you to look at case expressions or/and MultiWayIf as this may be a better way of writing this code:

health :: IO ()
health = do
  putStrLn "State your health using numbers 1 - 10: "
  str <- getLine
  case read str of
    num | num >= 1 && num <= 3 -> putStrLn "Your health is poor"
    num | num >= 4 && num <= 7 -> putStrLn "Your health is OK"
    num | num >= 8 && num <=10 -> putStrLn "your health is fanstastic"
    _ -> putStrLn "Wrong health range indicated"

answered May 17, 2019 at 12:27

Alexey Raga

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1 Comment

Daniel Wagner Over a year ago

Prefer readLn to getLine+read: it Does The Right Thing with exceptions when the input can't be parsed. In this particular case it doesn't matter, but it's a good habit to get into.

willeM_ Van Onsem · Accepted Answer · 2019-05-17 12:54:17Z

You make some mistakes here, mainly with the types:

The line:

num <- getLine

means that num is a String, but you later use it as a number. You probably want to use readLn :: Read a => IO a, and probably it is better to specify the type as well:

num <- readLn :: IO Int

In your first if cases:

if num >=1 && <=3

Is wrong since it is parsed as (num >= 1) && (<= 3). The right operator is thus not a Bool, but an (Num n, Ord n) => n -> Bool, so a function. A function is not True or False.

You should replace it with:

if num >=1 && num <=3

Later you write:

if num >=4 && getLine <=7

but here again the types do not match: getLine has type IO String, so it is not a number, it is even not an IO Int. You probably want to reuse the num, so:

if num >=4 && num <=7

Finally in the else stament, you write:

else "Wrong health range indicated"

But the type of health is IO (), not String, you thus should use putStrLn:

else putStrLn "Wrong health range indicated"

You can factor the putStrLn out, into:

health :: IO ()
health = do 
    putStrLn "State your health using numbers 1 - 10: "
    num <- readLn :: IO Int
    --putStrLn "Your health is: "
    putStrLn $
        if num <= 0 || num > 10 then "Wrong health range indicated"
        else if num <= 3 then "Your health is poor"
        else if num <= 7 then "Your health is OK"
        else "your health is fanstastic"

The above is still not ideal, since reading a string into an Int can go wrong. So it might help to use readMaybe here to make the program safer.

chi · Accepted Answer · 2019-05-17 12:22:00Z

4

To remove the first warning, replace tabs with spaces.

To fix the error, add num to <=3

if num >=1 && num <=3

You will also need to add a putStrLn here:

else putStrLn "Wrong health range indicated"

Also, you can't use getLine as in getLine <= 7. Shouldn't that be num instead? Do you really want to read another number from the user?

answered May 17, 2019 at 12:22

chi

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