Simple input function Haskell

Question

I want to write a simple function that reads a string from console and parses it to a custom data type.

My attempt:

data Custom = A | B Int | C String deriving Read

getFormula::IO Custom
getFormula = do 
    putStrLn("Introduce una formula: ")
    toParse <- getLine
    return read toParse::Custom

But this does not work and I do not know how to interpret the resulting compiling error. How do I fix this? What am I misunderstanding about how IO functions work?

EDIT: This is the error I get when I try to load the file into GCHI

test.hs:7:5:
Couldn't match type ‘String -> a0’ with ‘Custom’
Expected type: String -> Custom
  Actual type: String -> String -> a0
The function ‘return’ is applied to two arguments,
but its type ‘(String -> a0) -> String -> String -> a0’
has only three
In a stmt of a 'do' block: return read toParse :: Custom
In the expression:
  do { putStrLn ("Introduce una formula: ");
       toParse <- getLine;
         return read toParse :: Custom }

test.hs:7:5:
Couldn't match expected type ‘IO Custom’ with actual type ‘Custom’
In a stmt of a 'do' block: return read toParse :: Custom
In the expression:
  do { putStrLn ("Introduce una formula: ");
       toParse <- getLine;
         return read toParse :: Custom }
In an equation for ‘getFormula’:
    getFormula
      = do { putStrLn ("Introduce una formula: ");
             toParse <- getLine;
               return read toParse :: Custom }

You could use readLn directly instead getLine and read. — castletheperson
– castletheperson, Commented Dec 22, 2017 at 20:24

melpomene · Accepted Answer · 2017-12-22 19:50:39Z

You have two type errors. The second one is easier to understand:

getFormula::IO Custom

getFormula is declared to have type IO Custom.

    return read toParse::Custom

... but here you claim the last expression has type Custom. IO Custom is not the same as Custom, so the compiler complains.

By the way, your spacing is a bit weird. Why do you have :: jammed right against identifiers on the left and right?

    return read toParse :: Custom

looks less crowded and also less misleading: The :: Custom part applies to the whole expression on the left, not just a single variable.

The first error is a bit confused, but it contains a vital hint: The function ‘return’ is applied to two arguments.

return only takes one argument:

return read toParse

should be

return (read toParse)

In order to also fix the type annotation, you can use one of the following:

A bit clunky:
```
return (read toParse) :: IO Custom
```
Slightly neater (no need to specify IO):
```
return (read toParse :: Custom)
```
Easiest solution:
```
return (read toParse)
```
You don't need to explicitly specify a type here at all. The compiler already knows you're looking for a Custom because of the getFormula :: IO Custom declaration.

Fyodor Soikin · Accepted Answer · 2017-12-22 19:39:35Z

2

The return function has one argument, but you're giving it two - the first one is read, the second one is toParse.

Use parentheses to specify the application order:

return (read toParse :: Custom)

answered Dec 22, 2017 at 19:39

Fyodor Soikin

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