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752e4d5531
"Taught IDLE's autoident parser that "yield" is a keyword that begins a stmt. Along w/ the preceding change to keyword.py, making all this work w/ a future-stmt just looks harder and harder." --tim_one (From Rel 1.8: "Hack to make this still work with Python 1.5.2. ;-( " --fdrake)
589 lines
19 KiB
Python
589 lines
19 KiB
Python
import string
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import re
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import sys
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# Reason last stmt is continued (or C_NONE if it's not).
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C_NONE, C_BACKSLASH, C_STRING, C_BRACKET = range(4)
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if 0: # for throwaway debugging output
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def dump(*stuff):
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sys.__stdout__.write(string.join(map(str, stuff), " ") + "\n")
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# Find what looks like the start of a popular stmt.
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_synchre = re.compile(r"""
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^
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[ \t]*
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(?: if
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| for
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| while
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| else
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| def
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| return
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| assert
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| break
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| class
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| continue
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| elif
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| try
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| except
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| raise
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| import
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| yield
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)
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\b
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""", re.VERBOSE | re.MULTILINE).search
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# Match blank line or non-indenting comment line.
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_junkre = re.compile(r"""
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[ \t]*
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(?: \# \S .* )?
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\n
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""", re.VERBOSE).match
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# Match any flavor of string; the terminating quote is optional
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# so that we're robust in the face of incomplete program text.
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_match_stringre = re.compile(r"""
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\""" [^"\\]* (?:
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(?: \\. | "(?!"") )
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[^"\\]*
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)*
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(?: \""" )?
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| " [^"\\\n]* (?: \\. [^"\\\n]* )* "?
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| ''' [^'\\]* (?:
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(?: \\. | '(?!'') )
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[^'\\]*
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)*
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(?: ''' )?
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| ' [^'\\\n]* (?: \\. [^'\\\n]* )* '?
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""", re.VERBOSE | re.DOTALL).match
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# Match a line that starts with something interesting;
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# used to find the first item of a bracket structure.
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_itemre = re.compile(r"""
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[ \t]*
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[^\s#\\] # if we match, m.end()-1 is the interesting char
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""", re.VERBOSE).match
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# Match start of stmts that should be followed by a dedent.
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_closere = re.compile(r"""
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\s*
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(?: return
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| break
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| continue
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| raise
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| pass
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)
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\b
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""", re.VERBOSE).match
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# Chew up non-special chars as quickly as possible. If match is
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# successful, m.end() less 1 is the index of the last boring char
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# matched. If match is unsuccessful, the string starts with an
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# interesting char.
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_chew_ordinaryre = re.compile(r"""
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[^[\](){}#'"\\]+
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""", re.VERBOSE).match
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# Build translation table to map uninteresting chars to "x", open
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# brackets to "(", and close brackets to ")".
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_tran = ['x'] * 256
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for ch in "({[":
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_tran[ord(ch)] = '('
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for ch in ")}]":
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_tran[ord(ch)] = ')'
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for ch in "\"'\\\n#":
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_tran[ord(ch)] = ch
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_tran = string.join(_tran, '')
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del ch
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try:
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UnicodeType = type(unicode(""))
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except NameError:
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UnicodeType = None
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class Parser:
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def __init__(self, indentwidth, tabwidth):
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self.indentwidth = indentwidth
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self.tabwidth = tabwidth
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def set_str(self, str):
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assert len(str) == 0 or str[-1] == '\n'
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if type(str) is UnicodeType:
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# The parse functions have no idea what to do with Unicode, so
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# replace all Unicode characters with "x". This is "safe"
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# so long as the only characters germane to parsing the structure
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# of Python are 7-bit ASCII. It's *necessary* because Unicode
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# strings don't have a .translate() method that supports
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# deletechars.
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uniphooey = str
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str = []
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push = str.append
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for raw in map(ord, uniphooey):
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push(raw < 127 and chr(raw) or "x")
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str = "".join(str)
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self.str = str
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self.study_level = 0
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# Return index of a good place to begin parsing, as close to the
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# end of the string as possible. This will be the start of some
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# popular stmt like "if" or "def". Return None if none found:
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# the caller should pass more prior context then, if possible, or
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# if not (the entire program text up until the point of interest
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# has already been tried) pass 0 to set_lo.
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#
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# This will be reliable iff given a reliable is_char_in_string
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# function, meaning that when it says "no", it's absolutely
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# guaranteed that the char is not in a string.
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#
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# Ack, hack: in the shell window this kills us, because there's
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# no way to tell the differences between output, >>> etc and
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# user input. Indeed, IDLE's first output line makes the rest
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# look like it's in an unclosed paren!:
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# Python 1.5.2 (#0, Apr 13 1999, ...
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def find_good_parse_start(self, use_ps1, is_char_in_string=None,
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_rfind=string.rfind,
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_synchre=_synchre):
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str, pos = self.str, None
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if use_ps1:
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# shell window
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ps1 = '\n' + sys.ps1
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i = _rfind(str, ps1)
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if i >= 0:
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pos = i + len(ps1)
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# make it look like there's a newline instead
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# of ps1 at the start -- hacking here once avoids
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# repeated hackery later
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self.str = str[:pos-1] + '\n' + str[pos:]
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return pos
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# File window -- real work.
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if not is_char_in_string:
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# no clue -- make the caller pass everything
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return None
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# Peek back from the end for a good place to start,
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# but don't try too often; pos will be left None, or
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# bumped to a legitimate synch point.
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limit = len(str)
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for tries in range(5):
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i = _rfind(str, ":\n", 0, limit)
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if i < 0:
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break
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i = _rfind(str, '\n', 0, i) + 1 # start of colon line
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m = _synchre(str, i, limit)
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if m and not is_char_in_string(m.start()):
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pos = m.start()
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break
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limit = i
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if pos is None:
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# Nothing looks like a block-opener, or stuff does
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# but is_char_in_string keeps returning true; most likely
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# we're in or near a giant string, the colorizer hasn't
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# caught up enough to be helpful, or there simply *aren't*
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# any interesting stmts. In any of these cases we're
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# going to have to parse the whole thing to be sure, so
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# give it one last try from the start, but stop wasting
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# time here regardless of the outcome.
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m = _synchre(str)
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if m and not is_char_in_string(m.start()):
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pos = m.start()
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return pos
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# Peeking back worked; look forward until _synchre no longer
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# matches.
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i = pos + 1
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while 1:
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m = _synchre(str, i)
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if m:
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s, i = m.span()
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if not is_char_in_string(s):
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pos = s
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else:
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break
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return pos
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# Throw away the start of the string. Intended to be called with
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# find_good_parse_start's result.
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def set_lo(self, lo):
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assert lo == 0 or self.str[lo-1] == '\n'
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if lo > 0:
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self.str = self.str[lo:]
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# As quickly as humanly possible <wink>, find the line numbers (0-
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# based) of the non-continuation lines.
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# Creates self.{goodlines, continuation}.
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def _study1(self, _replace=string.replace, _find=string.find):
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if self.study_level >= 1:
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return
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self.study_level = 1
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# Map all uninteresting characters to "x", all open brackets
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# to "(", all close brackets to ")", then collapse runs of
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# uninteresting characters. This can cut the number of chars
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# by a factor of 10-40, and so greatly speed the following loop.
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str = self.str
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str = string.translate(str, _tran)
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str = _replace(str, 'xxxxxxxx', 'x')
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str = _replace(str, 'xxxx', 'x')
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str = _replace(str, 'xx', 'x')
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str = _replace(str, 'xx', 'x')
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str = _replace(str, '\nx', '\n')
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# note that replacing x\n with \n would be incorrect, because
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# x may be preceded by a backslash
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# March over the squashed version of the program, accumulating
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# the line numbers of non-continued stmts, and determining
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# whether & why the last stmt is a continuation.
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continuation = C_NONE
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level = lno = 0 # level is nesting level; lno is line number
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self.goodlines = goodlines = [0]
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push_good = goodlines.append
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i, n = 0, len(str)
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while i < n:
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ch = str[i]
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i = i+1
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# cases are checked in decreasing order of frequency
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if ch == 'x':
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continue
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if ch == '\n':
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lno = lno + 1
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if level == 0:
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push_good(lno)
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# else we're in an unclosed bracket structure
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continue
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if ch == '(':
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level = level + 1
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continue
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if ch == ')':
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if level:
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level = level - 1
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# else the program is invalid, but we can't complain
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continue
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if ch == '"' or ch == "'":
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# consume the string
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quote = ch
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if str[i-1:i+2] == quote * 3:
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quote = quote * 3
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w = len(quote) - 1
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i = i+w
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while i < n:
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ch = str[i]
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i = i+1
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if ch == 'x':
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continue
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if str[i-1:i+w] == quote:
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i = i+w
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break
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if ch == '\n':
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lno = lno + 1
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if w == 0:
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# unterminated single-quoted string
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if level == 0:
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push_good(lno)
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break
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continue
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if ch == '\\':
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assert i < n
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if str[i] == '\n':
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lno = lno + 1
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i = i+1
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continue
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# else comment char or paren inside string
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else:
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# didn't break out of the loop, so we're still
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# inside a string
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continuation = C_STRING
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continue # with outer loop
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if ch == '#':
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# consume the comment
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i = _find(str, '\n', i)
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assert i >= 0
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continue
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assert ch == '\\'
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assert i < n
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if str[i] == '\n':
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lno = lno + 1
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if i+1 == n:
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continuation = C_BACKSLASH
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i = i+1
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# The last stmt may be continued for all 3 reasons.
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# String continuation takes precedence over bracket
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# continuation, which beats backslash continuation.
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if continuation != C_STRING and level > 0:
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continuation = C_BRACKET
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self.continuation = continuation
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# Push the final line number as a sentinel value, regardless of
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# whether it's continued.
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assert (continuation == C_NONE) == (goodlines[-1] == lno)
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if goodlines[-1] != lno:
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push_good(lno)
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def get_continuation_type(self):
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self._study1()
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return self.continuation
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# study1 was sufficient to determine the continuation status,
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# but doing more requires looking at every character. study2
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# does this for the last interesting statement in the block.
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# Creates:
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# self.stmt_start, stmt_end
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# slice indices of last interesting stmt
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# self.lastch
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# last non-whitespace character before optional trailing
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# comment
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# self.lastopenbracketpos
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# if continuation is C_BRACKET, index of last open bracket
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def _study2(self, _rfind=string.rfind, _find=string.find,
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_ws=string.whitespace):
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if self.study_level >= 2:
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return
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self._study1()
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self.study_level = 2
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# Set p and q to slice indices of last interesting stmt.
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str, goodlines = self.str, self.goodlines
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i = len(goodlines) - 1
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p = len(str) # index of newest line
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while i:
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assert p
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# p is the index of the stmt at line number goodlines[i].
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# Move p back to the stmt at line number goodlines[i-1].
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q = p
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for nothing in range(goodlines[i-1], goodlines[i]):
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# tricky: sets p to 0 if no preceding newline
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p = _rfind(str, '\n', 0, p-1) + 1
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# The stmt str[p:q] isn't a continuation, but may be blank
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# or a non-indenting comment line.
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if _junkre(str, p):
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i = i-1
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else:
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break
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if i == 0:
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# nothing but junk!
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assert p == 0
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q = p
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self.stmt_start, self.stmt_end = p, q
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# Analyze this stmt, to find the last open bracket (if any)
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# and last interesting character (if any).
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lastch = ""
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stack = [] # stack of open bracket indices
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push_stack = stack.append
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while p < q:
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# suck up all except ()[]{}'"#\\
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m = _chew_ordinaryre(str, p, q)
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if m:
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# we skipped at least one boring char
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newp = m.end()
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# back up over totally boring whitespace
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i = newp - 1 # index of last boring char
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while i >= p and str[i] in " \t\n":
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i = i-1
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if i >= p:
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lastch = str[i]
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p = newp
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if p >= q:
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break
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ch = str[p]
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if ch in "([{":
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push_stack(p)
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lastch = ch
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p = p+1
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continue
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if ch in ")]}":
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if stack:
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del stack[-1]
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lastch = ch
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p = p+1
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continue
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if ch == '"' or ch == "'":
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# consume string
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# Note that study1 did this with a Python loop, but
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# we use a regexp here; the reason is speed in both
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# cases; the string may be huge, but study1 pre-squashed
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# strings to a couple of characters per line. study1
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# also needed to keep track of newlines, and we don't
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# have to.
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lastch = ch
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p = _match_stringre(str, p, q).end()
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continue
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if ch == '#':
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# consume comment and trailing newline
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p = _find(str, '\n', p, q) + 1
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assert p > 0
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continue
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assert ch == '\\'
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p = p+1 # beyond backslash
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assert p < q
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if str[p] != '\n':
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# the program is invalid, but can't complain
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lastch = ch + str[p]
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p = p+1 # beyond escaped char
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# end while p < q:
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self.lastch = lastch
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if stack:
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self.lastopenbracketpos = stack[-1]
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# Assuming continuation is C_BRACKET, return the number
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# of spaces the next line should be indented.
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def compute_bracket_indent(self, _find=string.find):
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self._study2()
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assert self.continuation == C_BRACKET
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j = self.lastopenbracketpos
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str = self.str
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n = len(str)
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origi = i = string.rfind(str, '\n', 0, j) + 1
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j = j+1 # one beyond open bracket
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# find first list item; set i to start of its line
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while j < n:
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m = _itemre(str, j)
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if m:
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j = m.end() - 1 # index of first interesting char
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extra = 0
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break
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else:
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# this line is junk; advance to next line
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i = j = _find(str, '\n', j) + 1
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else:
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# nothing interesting follows the bracket;
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# reproduce the bracket line's indentation + a level
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j = i = origi
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while str[j] in " \t":
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j = j+1
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extra = self.indentwidth
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return len(string.expandtabs(str[i:j],
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self.tabwidth)) + extra
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# Return number of physical lines in last stmt (whether or not
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# it's an interesting stmt! this is intended to be called when
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# continuation is C_BACKSLASH).
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def get_num_lines_in_stmt(self):
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self._study1()
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goodlines = self.goodlines
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return goodlines[-1] - goodlines[-2]
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# Assuming continuation is C_BACKSLASH, return the number of spaces
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# the next line should be indented. Also assuming the new line is
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# the first one following the initial line of the stmt.
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def compute_backslash_indent(self):
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self._study2()
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assert self.continuation == C_BACKSLASH
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str = self.str
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i = self.stmt_start
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while str[i] in " \t":
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i = i+1
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startpos = i
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# See whether the initial line starts an assignment stmt; i.e.,
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# look for an = operator
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endpos = string.find(str, '\n', startpos) + 1
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found = level = 0
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while i < endpos:
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ch = str[i]
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if ch in "([{":
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level = level + 1
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i = i+1
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elif ch in ")]}":
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if level:
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level = level - 1
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i = i+1
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elif ch == '"' or ch == "'":
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i = _match_stringre(str, i, endpos).end()
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elif ch == '#':
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break
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elif level == 0 and ch == '=' and \
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(i == 0 or str[i-1] not in "=<>!") and \
|
|
str[i+1] != '=':
|
|
found = 1
|
|
break
|
|
else:
|
|
i = i+1
|
|
|
|
if found:
|
|
# found a legit =, but it may be the last interesting
|
|
# thing on the line
|
|
i = i+1 # move beyond the =
|
|
found = re.match(r"\s*\\", str[i:endpos]) is None
|
|
|
|
if not found:
|
|
# oh well ... settle for moving beyond the first chunk
|
|
# of non-whitespace chars
|
|
i = startpos
|
|
while str[i] not in " \t\n":
|
|
i = i+1
|
|
|
|
return len(string.expandtabs(str[self.stmt_start :
|
|
i],
|
|
self.tabwidth)) + 1
|
|
|
|
# Return the leading whitespace on the initial line of the last
|
|
# interesting stmt.
|
|
|
|
def get_base_indent_string(self):
|
|
self._study2()
|
|
i, n = self.stmt_start, self.stmt_end
|
|
j = i
|
|
str = self.str
|
|
while j < n and str[j] in " \t":
|
|
j = j + 1
|
|
return str[i:j]
|
|
|
|
# Did the last interesting stmt open a block?
|
|
|
|
def is_block_opener(self):
|
|
self._study2()
|
|
return self.lastch == ':'
|
|
|
|
# Did the last interesting stmt close a block?
|
|
|
|
def is_block_closer(self):
|
|
self._study2()
|
|
return _closere(self.str, self.stmt_start) is not None
|
|
|
|
# index of last open bracket ({[, or None if none
|
|
lastopenbracketpos = None
|
|
|
|
def get_last_open_bracket_pos(self):
|
|
self._study2()
|
|
return self.lastopenbracketpos
|