HiveBrain v1.2.0
Get Started
← Back to all entries
patternModerate

Parsing strings with escaped characters using Parsec

Submitted by: @import:stackexchange-codereview··
0
Viewed 0 times
withescapedparsingusingcharactersparsecstrings

Problem

I have been working through the exercises in a Write Yourself a Scheme in 48 Hours/Parsing and hacked together something to get parseString to comprehend escaped characters. Also had some inspiration from Real World Haskell chapter 16 minus the applicative parts.

parseString :: Parser LispVal
parseString = do char '"'
                 x  noneOf "\""
          escaped = choice $ map tryEscaped escapedChars
          tryEscaped c = try $ char '\\' >> char (fst c) >> return (snd c)
          escapedChars = zip "bnfrt\\\"/" "\b\n\f\r\t\\\"/"


This works but I am not fond of my 'tryEscaped' definition. What are the alternatives?

Solution

Some suggestions:

  • Replace fst and snd with a pattern match or explicit function arguments.



  • Extract the common char '\\' parser. You can then avoid the try.



  • Strings with lots of escaped characters are hard to read and it's hard to visually


verify that the escape codes are correctly matched with their replacements. Consider
spelling them out and aligning them to make this easy to see.

Here's what I came up with:

escaped = char '\\' >> choice (zipWith escapedChar codes replacements)
escapedChar code replacement = char code >> return replacement
codes        = ['b',  'n',  'f',  'r',  't',  '\\', '\"', '/']
replacements = ['\b', '\n', '\f', '\r', '\t', '\\', '\"', '/']

Code Snippets

escaped = char '\\' >> choice (zipWith escapedChar codes replacements)
escapedChar code replacement = char code >> return replacement
codes        = ['b',  'n',  'f',  'r',  't',  '\\', '\"', '/']
replacements = ['\b', '\n', '\f', '\r', '\t', '\\', '\"', '/']

Context

StackExchange Code Review Q#2406, answer score: 11

Revisions (0)

No revisions yet.