Haskell: recursively convert hex string to integer?

haskell, hex, recursion, type-conversion

Solution

You are really close. your error is on this line:

    | length hxStr /= 0 = (hexChar(last(hxStr)))+(10*parseHex(init(hxStr)))

Think about why you are multiplying by 10. Remember ... Hexadecimal is base 16.

Problem

For my homework assignment, I need to convert a hexadecimal string to a base-10 integer using a recursive function (with as many helper methods as necessary). This is what I've got so far: ``` -- Question 1, part (c): hexChar :: Char -> Integer hexChar ch | ch == '0' = 0 | ch == '1' = 1 | ch == '2' = 2 | ch == '3' = 3 | ch == '4' = 4 | ch == '5' = 5 | ch == '6' = 6 | ch == '7' = 7 | ch == '8' = 8 | ch == '9' = 9 | ch == 'A' = 10 | ch == 'B' = 11 | ch == 'C' = 12 | ch == 'D' = 13 | ch == 'E' = 14 | ch == 'F' = 15 | otherwise = 0 parseHex :: String -> Integer parseHex hxStr | length hxStr /= 0 = (hexChar(last(hxStr)))+(10*parseHex(init(hxStr))) | otherwise = 0 ``` However, this does not produce the correct results. Does anyone know of the correct way to do this?

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