changegroup: simplify by not reusing 'prog(ress)' instance
Just create a new instance of the 'prog' class for each step instead
of replacing its fields and resetting the counter.
# parser.py - simple top-down operator precedence parser for mercurial
#
# Copyright 2010 Matt Mackall <mpm@selenic.com>
#
# This software may be used and distributed according to the terms of the
# GNU General Public License version 2 or any later version.
# see http://effbot.org/zone/simple-top-down-parsing.htm and
# http://eli.thegreenplace.net/2010/01/02/top-down-operator-precedence-parsing/
# for background
# takes a tokenizer and elements
# tokenizer is an iterator that returns type, value pairs
# elements is a mapping of types to binding strength, prefix and infix actions
# an action is a tree node name, a tree label, and an optional match
# __call__(program) parses program into a labeled tree
import error
from i18n import _
class parser(object):
def __init__(self, tokenizer, elements, methods=None):
self._tokenizer = tokenizer
self._elements = elements
self._methods = methods
self.current = None
def _advance(self):
'advance the tokenizer'
t = self.current
self.current = next(self._iter, None)
return t
def _match(self, m, pos):
'make sure the tokenizer matches an end condition'
if self.current[0] != m:
raise error.ParseError(_("unexpected token: %s") % self.current[0],
self.current[2])
self._advance()
def _parse(self, bind=0):
token, value, pos = self._advance()
# handle prefix rules on current token
prefix = self._elements[token][1]
if not prefix:
raise error.ParseError(_("not a prefix: %s") % token, pos)
if len(prefix) == 1:
expr = (prefix[0], value)
else:
if len(prefix) > 2 and prefix[2] == self.current[0]:
self._match(prefix[2], pos)
expr = (prefix[0], None)
else:
expr = (prefix[0], self._parse(prefix[1]))
if len(prefix) > 2:
self._match(prefix[2], pos)
# gather tokens until we meet a lower binding strength
while bind < self._elements[self.current[0]][0]:
token, value, pos = self._advance()
e = self._elements[token]
# check for suffix - next token isn't a valid prefix
if len(e) == 4 and not self._elements[self.current[0]][1]:
suffix = e[3]
expr = (suffix[0], expr)
else:
# handle infix rules
if len(e) < 3 or not e[2]:
raise error.ParseError(_("not an infix: %s") % token, pos)
infix = e[2]
if len(infix) == 3 and infix[2] == self.current[0]:
self._match(infix[2], pos)
expr = (infix[0], expr, (None))
else:
expr = (infix[0], expr, self._parse(infix[1]))
if len(infix) == 3:
self._match(infix[2], pos)
return expr
def parse(self, message, lookup=None):
'generate a parse tree from a message'
if lookup:
self._iter = self._tokenizer(message, lookup)
else:
self._iter = self._tokenizer(message)
self._advance()
res = self._parse()
token, value, pos = self.current
return res, pos
def eval(self, tree):
'recursively evaluate a parse tree using node methods'
if not isinstance(tree, tuple):
return tree
return self._methods[tree[0]](*[self.eval(t) for t in tree[1:]])
def __call__(self, message):
'parse a message into a parse tree and evaluate if methods given'
t = self.parse(message)
if self._methods:
return self.eval(t)
return t
def _prettyformat(tree, leafnodes, level, lines):
if not isinstance(tree, tuple) or tree[0] in leafnodes:
lines.append((level, str(tree)))
else:
lines.append((level, '(%s' % tree[0]))
for s in tree[1:]:
_prettyformat(s, leafnodes, level + 1, lines)
lines[-1:] = [(lines[-1][0], lines[-1][1] + ')')]
def prettyformat(tree, leafnodes):
lines = []
_prettyformat(tree, leafnodes, 0, lines)
output = '\n'.join((' ' * l + s) for l, s in lines)
return output
def simplifyinfixops(tree, targetnodes):
"""Flatten chained infix operations to reduce usage of Python stack
>>> def f(tree):
... print prettyformat(simplifyinfixops(tree, ('or',)), ('symbol',))
>>> f(('or',
... ('or',
... ('symbol', '1'),
... ('symbol', '2')),
... ('symbol', '3')))
(or
('symbol', '1')
('symbol', '2')
('symbol', '3'))
>>> f(('func',
... ('symbol', 'p1'),
... ('or',
... ('or',
... ('func',
... ('symbol', 'sort'),
... ('list',
... ('or',
... ('or',
... ('symbol', '1'),
... ('symbol', '2')),
... ('symbol', '3')),
... ('negate',
... ('symbol', 'rev')))),
... ('and',
... ('symbol', '4'),
... ('group',
... ('or',
... ('or',
... ('symbol', '5'),
... ('symbol', '6')),
... ('symbol', '7'))))),
... ('symbol', '8'))))
(func
('symbol', 'p1')
(or
(func
('symbol', 'sort')
(list
(or
('symbol', '1')
('symbol', '2')
('symbol', '3'))
(negate
('symbol', 'rev'))))
(and
('symbol', '4')
(group
(or
('symbol', '5')
('symbol', '6')
('symbol', '7'))))
('symbol', '8')))
"""
if not isinstance(tree, tuple):
return tree
op = tree[0]
if op not in targetnodes:
return (op,) + tuple(simplifyinfixops(x, targetnodes) for x in tree[1:])
# walk down left nodes taking each right node. no recursion to left nodes
# because infix operators are left-associative, i.e. left tree is deep.
# e.g. '1 + 2 + 3' -> (+ (+ 1 2) 3) -> (+ 1 2 3)
simplified = []
x = tree
while x[0] == op:
l, r = x[1:]
simplified.append(simplifyinfixops(r, targetnodes))
x = l
simplified.append(simplifyinfixops(x, targetnodes))
simplified.append(op)
return tuple(reversed(simplified))