CODEWITHSUNDEEP
haskellfold
BrownE
BrownE

Reputation: 1719

Foldr issues (Haskell)

I'm trying to convert from a table which is of type ([Char, Int]) to a string tab2str :: Table -> String (that follows some specific formatting patterns.)

I'm using foldr (as the title implies) but I'm having some issues getting the exact function to work - i.e. It errors. My function looks like this:

tab2str xs = foldr (++) ' ' $ map (\char count -> show char ++ ':' ++ show count ++ '\n') xs

The output should be each letter in the table, a colon, then \n. So a test might be:

tab2str test1 == "F: 1\no: 1\nl: 1\nd: 1\nr: 1\n"

where test1 == [(F, 1), (o, 1), (l, 1), (d, 1), (r, 1)]

Any help gratefully received.

Upvotes: 1

Views: 265

Answers (1)

ulidtko
ulidtko

Reputation: 15652

After minimal corrections this typechecks:

tab2str xs = foldr (++) " " $ map (\(char, count) -> show char ++ ":" ++ show count ++ "\n") xs

– but produces not quite what you want.

You'll probably like this better:

tab2str table = concat $ map formatRow table
    where formatRow (char, count) = [char] ++ ": " ++ show count ++ "\n"

Then your test example:

ghci> let test1 = [('F', 1), ('o', 1), ('l', 1), ('d', 1), ('r', 1)]
ghci> tab2str test1
"F: 1\no: 1\nl: 1\nd: 1\nr: 1\n"
ghci> putStr $ tab2str test1
F: 1
o: 1
l: 1
d: 1
r: 1
ghci>

Upvotes: 1

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