Get a string output using Rply throws ParserGeneratorError: Expecting :

Question

Trying to implement parser using rply its just like ply

#Input = ['ABC']
lg = LexerGenerator()
lg.add('String', r'\D')

l = lg.build()
for token in l.lex('ABC'):
  print(token)

Generated Lexer by above code

class String(BaseBox):
  def __init__(self, value):
    self.value = value 
  def eval(self):
    return self.value


from rply import ParserGenerator
pg = ParserGenerator(
# A list of all token names, accepted by the parser. 
     ['String']
)
@pg.production('program: String')
def program(p):
    return p[0].value

parser = pg.build() #should parse string 'ABC'

ParserGeneratorError: Expecting :

Confused this isn't even in the documentation, please reply. Want my output to read the stringn as 'ABC'

Answer 1

Your code indicates that you tested the lexer. However, the test indicates that the lexer is not producing the correct tokens:

>>> for token in l.lex('ABC'):
...   print(token)
... 
Token('String', 'A')
Token('String', 'B')
Token('String', 'C')

The expected output would have been

Token('String', 'ABC')

The reason you're splitting on individual characters is that your pattern for recognising String only matches a single character:

lg.add('String', r'\D')

Probably you wanted something more like

lg.add('String', r'\D+')

But note that \D matches anything which is not a decimal digit, including whitespace, punctuation and control characters. Perhaps that is what you wanted, but it seems a bit too permissive to me.

---

Unrelated, but you'll have to fix this as well: as far as I know, RPLY (like Ply) requires you to write grammar rules with spaces around the colon, so your parser function needs to be

@pg.production('program : String')
def program(p):
    return p[0].value