This file is indexed.

/usr/lib/python2.7/idlelib/PyParse.py is in idle-python2.7 2.7.3-0ubuntu3.

This file is owned by root:root, with mode 0o644.

The actual contents of the file can be viewed below.

  1
  2
  3
  4
  5
  6
  7
  8
  9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 41
 42
 43
 44
 45
 46
 47
 48
 49
 50
 51
 52
 53
 54
 55
 56
 57
 58
 59
 60
 61
 62
 63
 64
 65
 66
 67
 68
 69
 70
 71
 72
 73
 74
 75
 76
 77
 78
 79
 80
 81
 82
 83
 84
 85
 86
 87
 88
 89
 90
 91
 92
 93
 94
 95
 96
 97
 98
 99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
import re
import sys

# Reason last stmt is continued (or C_NONE if it's not).
(C_NONE, C_BACKSLASH, C_STRING_FIRST_LINE,
 C_STRING_NEXT_LINES, C_BRACKET) = range(5)

if 0:   # for throwaway debugging output
    def dump(*stuff):
        sys.__stdout__.write(" ".join(map(str, stuff)) + "\n")

# Find what looks like the start of a popular stmt.

_synchre = re.compile(r"""
    ^
    [ \t]*
    (?: while
    |   else
    |   def
    |   return
    |   assert
    |   break
    |   class
    |   continue
    |   elif
    |   try
    |   except
    |   raise
    |   import
    |   yield
    )
    \b
""", re.VERBOSE | re.MULTILINE).search

# Match blank line or non-indenting comment line.

_junkre = re.compile(r"""
    [ \t]*
    (?: \# \S .* )?
    \n
""", re.VERBOSE).match

# Match any flavor of string; the terminating quote is optional
# so that we're robust in the face of incomplete program text.

_match_stringre = re.compile(r"""
    \""" [^"\\]* (?:
                     (?: \\. | "(?!"") )
                     [^"\\]*
                 )*
    (?: \""" )?

|   " [^"\\\n]* (?: \\. [^"\\\n]* )* "?

|   ''' [^'\\]* (?:
                   (?: \\. | '(?!'') )
                   [^'\\]*
                )*
    (?: ''' )?

|   ' [^'\\\n]* (?: \\. [^'\\\n]* )* '?
""", re.VERBOSE | re.DOTALL).match

# Match a line that starts with something interesting;
# used to find the first item of a bracket structure.

_itemre = re.compile(r"""
    [ \t]*
    [^\s#\\]    # if we match, m.end()-1 is the interesting char
""", re.VERBOSE).match

# Match start of stmts that should be followed by a dedent.

_closere = re.compile(r"""
    \s*
    (?: return
    |   break
    |   continue
    |   raise
    |   pass
    )
    \b
""", re.VERBOSE).match

# Chew up non-special chars as quickly as possible.  If match is
# successful, m.end() less 1 is the index of the last boring char
# matched.  If match is unsuccessful, the string starts with an
# interesting char.

_chew_ordinaryre = re.compile(r"""
    [^[\](){}#'"\\]+
""", re.VERBOSE).match

# Build translation table to map uninteresting chars to "x", open
# brackets to "(", and close brackets to ")".

_tran = ['x'] * 256
for ch in "({[":
    _tran[ord(ch)] = '('
for ch in ")}]":
    _tran[ord(ch)] = ')'
for ch in "\"'\\\n#":
    _tran[ord(ch)] = ch
_tran = ''.join(_tran)
del ch

try:
    UnicodeType = type(unicode(""))
except NameError:
    UnicodeType = None

class Parser:

    def __init__(self, indentwidth, tabwidth):
        self.indentwidth = indentwidth
        self.tabwidth = tabwidth

    def set_str(self, str):
        assert len(str) == 0 or str[-1] == '\n'
        if type(str) is UnicodeType:
            # The parse functions have no idea what to do with Unicode, so
            # replace all Unicode characters with "x".  This is "safe"
            # so long as the only characters germane to parsing the structure
            # of Python are 7-bit ASCII.  It's *necessary* because Unicode
            # strings don't have a .translate() method that supports
            # deletechars.
            uniphooey = str
            str = []
            push = str.append
            for raw in map(ord, uniphooey):
                push(raw < 127 and chr(raw) or "x")
            str = "".join(str)
        self.str = str
        self.study_level = 0

    # Return index of a good place to begin parsing, as close to the
    # end of the string as possible.  This will be the start of some
    # popular stmt like "if" or "def".  Return None if none found:
    # the caller should pass more prior context then, if possible, or
    # if not (the entire program text up until the point of interest
    # has already been tried) pass 0 to set_lo.
    #
    # This will be reliable iff given a reliable is_char_in_string
    # function, meaning that when it says "no", it's absolutely
    # guaranteed that the char is not in a string.

    def find_good_parse_start(self, is_char_in_string=None,
                              _synchre=_synchre):
        str, pos = self.str, None

        if not is_char_in_string:
            # no clue -- make the caller pass everything
            return None

        # Peek back from the end for a good place to start,
        # but don't try too often; pos will be left None, or
        # bumped to a legitimate synch point.
        limit = len(str)
        for tries in range(5):
            i = str.rfind(":\n", 0, limit)
            if i < 0:
                break
            i = str.rfind('\n', 0, i) + 1  # start of colon line
            m = _synchre(str, i, limit)
            if m and not is_char_in_string(m.start()):
                pos = m.start()
                break
            limit = i
        if pos is None:
            # Nothing looks like a block-opener, or stuff does
            # but is_char_in_string keeps returning true; most likely
            # we're in or near a giant string, the colorizer hasn't
            # caught up enough to be helpful, or there simply *aren't*
            # any interesting stmts.  In any of these cases we're
            # going to have to parse the whole thing to be sure, so
            # give it one last try from the start, but stop wasting
            # time here regardless of the outcome.
            m = _synchre(str)
            if m and not is_char_in_string(m.start()):
                pos = m.start()
            return pos

        # Peeking back worked; look forward until _synchre no longer
        # matches.
        i = pos + 1
        while 1:
            m = _synchre(str, i)
            if m:
                s, i = m.span()
                if not is_char_in_string(s):
                    pos = s
            else:
                break
        return pos

    # Throw away the start of the string.  Intended to be called with
    # find_good_parse_start's result.

    def set_lo(self, lo):
        assert lo == 0 or self.str[lo-1] == '\n'
        if lo > 0:
            self.str = self.str[lo:]

    # As quickly as humanly possible <wink>, find the line numbers (0-
    # based) of the non-continuation lines.
    # Creates self.{goodlines, continuation}.

    def _study1(self):
        if self.study_level >= 1:
            return
        self.study_level = 1

        # Map all uninteresting characters to "x", all open brackets
        # to "(", all close brackets to ")", then collapse runs of
        # uninteresting characters.  This can cut the number of chars
        # by a factor of 10-40, and so greatly speed the following loop.
        str = self.str
        str = str.translate(_tran)
        str = str.replace('xxxxxxxx', 'x')
        str = str.replace('xxxx', 'x')
        str = str.replace('xx', 'x')
        str = str.replace('xx', 'x')
        str = str.replace('\nx', '\n')
        # note that replacing x\n with \n would be incorrect, because
        # x may be preceded by a backslash

        # March over the squashed version of the program, accumulating
        # the line numbers of non-continued stmts, and determining
        # whether & why the last stmt is a continuation.
        continuation = C_NONE
        level = lno = 0     # level is nesting level; lno is line number
        self.goodlines = goodlines = [0]
        push_good = goodlines.append
        i, n = 0, len(str)
        while i < n:
            ch = str[i]
            i = i+1

            # cases are checked in decreasing order of frequency
            if ch == 'x':
                continue

            if ch == '\n':
                lno = lno + 1
                if level == 0:
                    push_good(lno)
                    # else we're in an unclosed bracket structure
                continue

            if ch == '(':
                level = level + 1
                continue

            if ch == ')':
                if level:
                    level = level - 1
                    # else the program is invalid, but we can't complain
                continue

            if ch == '"' or ch == "'":
                # consume the string
                quote = ch
                if str[i-1:i+2] == quote * 3:
                    quote = quote * 3
                firstlno = lno
                w = len(quote) - 1
                i = i+w
                while i < n:
                    ch = str[i]
                    i = i+1

                    if ch == 'x':
                        continue

                    if str[i-1:i+w] == quote:
                        i = i+w
                        break

                    if ch == '\n':
                        lno = lno + 1
                        if w == 0:
                            # unterminated single-quoted string
                            if level == 0:
                                push_good(lno)
                            break
                        continue

                    if ch == '\\':
                        assert i < n
                        if str[i] == '\n':
                            lno = lno + 1
                        i = i+1
                        continue

                    # else comment char or paren inside string

                else:
                    # didn't break out of the loop, so we're still
                    # inside a string
                    if (lno - 1) == firstlno:
                        # before the previous \n in str, we were in the first
                        # line of the string
                        continuation = C_STRING_FIRST_LINE
                    else:
                        continuation = C_STRING_NEXT_LINES
                continue    # with outer loop

            if ch == '#':
                # consume the comment
                i = str.find('\n', i)
                assert i >= 0
                continue

            assert ch == '\\'
            assert i < n
            if str[i] == '\n':
                lno = lno + 1
                if i+1 == n:
                    continuation = C_BACKSLASH
            i = i+1

        # The last stmt may be continued for all 3 reasons.
        # String continuation takes precedence over bracket
        # continuation, which beats backslash continuation.
        if (continuation != C_STRING_FIRST_LINE
            and continuation != C_STRING_NEXT_LINES and level > 0):
            continuation = C_BRACKET
        self.continuation = continuation

        # Push the final line number as a sentinel value, regardless of
        # whether it's continued.
        assert (continuation == C_NONE) == (goodlines[-1] == lno)
        if goodlines[-1] != lno:
            push_good(lno)

    def get_continuation_type(self):
        self._study1()
        return self.continuation

    # study1 was sufficient to determine the continuation status,
    # but doing more requires looking at every character.  study2
    # does this for the last interesting statement in the block.
    # Creates:
    #     self.stmt_start, stmt_end
    #         slice indices of last interesting stmt
    #     self.stmt_bracketing
    #         the bracketing structure of the last interesting stmt;
    #         for example, for the statement "say(boo) or die", stmt_bracketing
    #         will be [(0, 0), (3, 1), (8, 0)]. Strings and comments are
    #         treated as brackets, for the matter.
    #     self.lastch
    #         last non-whitespace character before optional trailing
    #         comment
    #     self.lastopenbracketpos
    #         if continuation is C_BRACKET, index of last open bracket

    def _study2(self):
        if self.study_level >= 2:
            return
        self._study1()
        self.study_level = 2

        # Set p and q to slice indices of last interesting stmt.
        str, goodlines = self.str, self.goodlines
        i = len(goodlines) - 1
        p = len(str)    # index of newest line
        while i:
            assert p
            # p is the index of the stmt at line number goodlines[i].
            # Move p back to the stmt at line number goodlines[i-1].
            q = p
            for nothing in range(goodlines[i-1], goodlines[i]):
                # tricky: sets p to 0 if no preceding newline
                p = str.rfind('\n', 0, p-1) + 1
            # The stmt str[p:q] isn't a continuation, but may be blank
            # or a non-indenting comment line.
            if  _junkre(str, p):
                i = i-1
            else:
                break
        if i == 0:
            # nothing but junk!
            assert p == 0
            q = p
        self.stmt_start, self.stmt_end = p, q

        # Analyze this stmt, to find the last open bracket (if any)
        # and last interesting character (if any).
        lastch = ""
        stack = []  # stack of open bracket indices
        push_stack = stack.append
        bracketing = [(p, 0)]
        while p < q:
            # suck up all except ()[]{}'"#\\
            m = _chew_ordinaryre(str, p, q)
            if m:
                # we skipped at least one boring char
                newp = m.end()
                # back up over totally boring whitespace
                i = newp - 1    # index of last boring char
                while i >= p and str[i] in " \t\n":
                    i = i-1
                if i >= p:
                    lastch = str[i]
                p = newp
                if p >= q:
                    break

            ch = str[p]

            if ch in "([{":
                push_stack(p)
                bracketing.append((p, len(stack)))
                lastch = ch
                p = p+1
                continue

            if ch in ")]}":
                if stack:
                    del stack[-1]
                lastch = ch
                p = p+1
                bracketing.append((p, len(stack)))
                continue

            if ch == '"' or ch == "'":
                # consume string
                # Note that study1 did this with a Python loop, but
                # we use a regexp here; the reason is speed in both
                # cases; the string may be huge, but study1 pre-squashed
                # strings to a couple of characters per line.  study1
                # also needed to keep track of newlines, and we don't
                # have to.
                bracketing.append((p, len(stack)+1))
                lastch = ch
                p = _match_stringre(str, p, q).end()
                bracketing.append((p, len(stack)))
                continue

            if ch == '#':
                # consume comment and trailing newline
                bracketing.append((p, len(stack)+1))
                p = str.find('\n', p, q) + 1
                assert p > 0
                bracketing.append((p, len(stack)))
                continue

            assert ch == '\\'
            p = p+1     # beyond backslash
            assert p < q
            if str[p] != '\n':
                # the program is invalid, but can't complain
                lastch = ch + str[p]
            p = p+1     # beyond escaped char

        # end while p < q:

        self.lastch = lastch
        if stack:
            self.lastopenbracketpos = stack[-1]
        self.stmt_bracketing = tuple(bracketing)

    # Assuming continuation is C_BRACKET, return the number
    # of spaces the next line should be indented.

    def compute_bracket_indent(self):
        self._study2()
        assert self.continuation == C_BRACKET
        j = self.lastopenbracketpos
        str = self.str
        n = len(str)
        origi = i = str.rfind('\n', 0, j) + 1
        j = j+1     # one beyond open bracket
        # find first list item; set i to start of its line
        while j < n:
            m = _itemre(str, j)
            if m:
                j = m.end() - 1     # index of first interesting char
                extra = 0
                break
            else:
                # this line is junk; advance to next line
                i = j = str.find('\n', j) + 1
        else:
            # nothing interesting follows the bracket;
            # reproduce the bracket line's indentation + a level
            j = i = origi
            while str[j] in " \t":
                j = j+1
            extra = self.indentwidth
        return len(str[i:j].expandtabs(self.tabwidth)) + extra

    # Return number of physical lines in last stmt (whether or not
    # it's an interesting stmt!  this is intended to be called when
    # continuation is C_BACKSLASH).

    def get_num_lines_in_stmt(self):
        self._study1()
        goodlines = self.goodlines
        return goodlines[-1] - goodlines[-2]

    # Assuming continuation is C_BACKSLASH, return the number of spaces
    # the next line should be indented.  Also assuming the new line is
    # the first one following the initial line of the stmt.

    def compute_backslash_indent(self):
        self._study2()
        assert self.continuation == C_BACKSLASH
        str = self.str
        i = self.stmt_start
        while str[i] in " \t":
            i = i+1
        startpos = i

        # See whether the initial line starts an assignment stmt; i.e.,
        # look for an = operator
        endpos = str.find('\n', startpos) + 1
        found = level = 0
        while i < endpos:
            ch = str[i]
            if ch in "([{":
                level = level + 1
                i = i+1
            elif ch in ")]}":
                if level:
                    level = level - 1
                i = i+1
            elif ch == '"' or ch == "'":
                i = _match_stringre(str, i, endpos).end()
            elif ch == '#':
                break
            elif level == 0 and ch == '=' and \
                   (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(str[self.stmt_start:i].expandtabs(\
                                     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

    # the structure of the bracketing of the last interesting statement,
    # in the format defined in _study2, or None if the text didn't contain
    # anything
    stmt_bracketing = None

    def get_last_stmt_bracketing(self):
        self._study2()
        return self.stmt_bracketing