Haskell + output String as type String

Question

I've written a basic recursive function:

bibliography_rec :: [(String, String, Int)] -> String
bibliography_rec [] = ""
bibliography_rec (x:xs) = (citeBook x) ++ "\n" ++ (bibliography_rec xs)

citeBook simply reformats the tuple into a String.

When run with this input:

ghci> bibliography_rec [("Herman Melville", "Moby Dick", 1851),("Georgy Poo", "Alex Janakos", 1666)]

It produces:

"Moby Dick (Herman Melville, 1851)\nAlex Janakos (Georgy Poo, 1666)\n"

I need line by line printing so I used this:

bibliography_rec (x:xs) = putStr ((citeBook x) ++ "\n" ++ (bibliography_rec xs))

My problem is my output NEEDS to be of type String NOT IO ()

I've been stuck on this for way too long so any help is great!

Values of type String cannot produce output. So your requirements ("I must produce output" and "the return type must be String") are in conflict. — Daniel Wagner
– Daniel Wagner, Commented Jan 11, 2016 at 21:19

luqui · Accepted Answer · 2016-01-11 20:25:35Z

5

Looks like you're already there, you just need to putStrLn the string instead of printing it (which is what ghci does by default). print runs its argument through show first, so it will quote the escape characters like "\n".

ghci> putStrLn $ bibliography_rec [...]

answered Jan 11, 2016 at 20:25

luqui

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3 Comments

ajanakos Over a year ago

Only problem is I need to generate the String with only ghci> bibliography_rec [(...),(...)] or is there something im still missing. Also what does the $ do??

bheklilr Over a year ago

@ajanakos There's definitely something you're missing. A String value is just a String value, in Haskell that means that it has no side effects and simply can't do things like print a value to the screen. In GHCi when you evaluate a value, it performs an IO action behind the scenes so that you get to see a representation of that value displayed, since otherwise the "interactive" part of GHCi wouldn't really be useful. Note that code entered into GHCi is not the same as executing a .hs file.

bheklilr Over a year ago

@ajanakos The $ operator has many, many discussions, clarifications, etc all over stackoverflow and the rest of the internet, I won't go into detail here, but suffice to say that it's a do-nothing operator with a specific order of operations so that you can drop parentheses frequently. The above line of code could have been equivalently written as putStrLn (bibliography_rec [...]) and it would have been identical in execution. This comes in handy with things like f $ g $ h 1 instead of f (g (h 1))), where the parens mostly just add noise.

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