Current File : //proc/thread-self/root/opt/alt/python34/lib64/python3.4/__pycache__/pickletools.cpython-34.pyo
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Extensive comments about the pickle protocols and pickle-machine opcodes
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genops(pickle)
Generate all the opcodes in a pickle, as (opcode, arg, position) triples.
dis(pickle, out=None, memo=None, indentlevel=4)
Print a symbolic disassembly of a pickle.
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>>> import io
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255
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65535
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>>> import io
>>> read_int4(io.BytesIO(b'\xff\x00\x00\x00'))
255
>>> read_int4(io.BytesIO(b'\x00\x00\x00\x80')) == -(2**31)
True
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>>> import io
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255
>>> read_uint4(io.BytesIO(b'\x00\x00\x00\x80')) == 2**31
True
rz<Irz'not enough data in stream to read uint4N)rrrr)rrrrr�
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>>> import io
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255
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True
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>>> import io
>>> read_stringnl(io.BytesIO(b"'abcd'\nefg\n"))
'abcd'
>>> read_stringnl(io.BytesIO(b"\n"))
Traceback (most recent call last):
...
ValueError: no string quotes around b''
>>> read_stringnl(io.BytesIO(b"\n"), stripquotes=False)
''
>>> read_stringnl(io.BytesIO(b"''\n"))
''
>>> read_stringnl(io.BytesIO(b'"abcd"'))
Traceback (most recent call last):
...
ValueError: no newline found when trying to read stringnl
Embedded escapes are undone in the result.
>>> read_stringnl(io.BytesIO(br"'a\n\\b\x00c\td'" + b"\n'e'"))
'a\n\\b\x00c\td'
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>>> import io
>>> read_stringnl_noescape_pair(io.BytesIO(b"Queue\nEmpty\njunk"))
'Queue Empty'
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a single blank separating the two strings.
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>>> import io
>>> read_string1(io.BytesIO(b"\x00"))
''
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'abc'
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r;�string1z�A counted string.
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>>> import io
>>> read_string4(io.BytesIO(b"\x00\x00\x00\x00abc"))
''
>>> read_string4(io.BytesIO(b"\x03\x00\x00\x00abcdef"))
'abc'
>>> read_string4(io.BytesIO(b"\x00\x00\x00\x03abcdef"))
Traceback (most recent call last):
...
ValueError: expected 50331648 bytes in a string4, but only 6 remain
rzstring4 byte count < 0: %dzlatin-1z2expected %d bytes in a string4, but only %d remainN)r"rrrr2)rrrrrr�read_string4�s
r=�string4z�A counted string.
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>>> import io
>>> read_bytes1(io.BytesIO(b"\x00"))
b''
>>> read_bytes1(io.BytesIO(b"\x03abcdef"))
b'abc'
z1expected %d bytes in a bytes1, but only %d remainN)rrrr)rrrrrr�read_bytes1�s r?�bytes1z�A counted bytes string.
The first argument is a 1-byte unsigned int giving the number
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>>> import io
>>> read_bytes1(io.BytesIO(b"\x00"))
b''
>>> read_bytes1(io.BytesIO(b"\x03abcdef"))
b'abc'
z1expected %d bytes in a bytes1, but only %d remainN)rrrr)rrrrrrr?�s z�A counted bytes string.
The first argument is a 1-byte unsigned int giving the number
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cCsst|�}|tjkr.td|��n|j|�}t|�|krS|Std|t|�f��dS)aN
>>> import io
>>> read_bytes4(io.BytesIO(b"\x00\x00\x00\x00abc"))
b''
>>> read_bytes4(io.BytesIO(b"\x03\x00\x00\x00abcdef"))
b'abc'
>>> read_bytes4(io.BytesIO(b"\x00\x00\x00\x03abcdef"))
Traceback (most recent call last):
...
ValueError: expected 50331648 bytes in a bytes4, but only 6 remain
z#bytes4 byte count > sys.maxsize: %dz1expected %d bytes in a bytes4, but only %d remainN)r$�sys�maxsizerrr)rrrrrr�read_bytes4s
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>>> import io, struct, sys
>>> read_bytes8(io.BytesIO(b"\x00\x00\x00\x00\x00\x00\x00\x00abc"))
b''
>>> read_bytes8(io.BytesIO(b"\x03\x00\x00\x00\x00\x00\x00\x00abcdef"))
b'abc'
>>> bigsize8 = struct.pack("<Q", sys.maxsize//3)
>>> read_bytes8(io.BytesIO(bigsize8 + b"abcdef")) #doctest: +ELLIPSIS
Traceback (most recent call last):
...
ValueError: expected ... bytes in a bytes8, but only 6 remain
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The first argument is a 8-byte little-endian unsigned int giving
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>>> import io
>>> read_unicodestringnl(io.BytesIO(b"abc\\uabcd\njunk")) == 'abc\uabcd'
True
s
z4no newline found when trying to read unicodestringnlNrzraw-unicode-escaper,)r-r.r�str)rrrrr�read_unicodestringnlUs
rH�unicodestringnlz�A newline-terminated Unicode string.
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>>> import io
>>> s = 'abcd\uabcd'
>>> enc = s.encode('utf-8')
>>> enc
b'abcd\xea\xaf\x8d'
>>> n = bytes([len(enc)]) # little-endian 1-byte length
>>> t = read_unicodestring1(io.BytesIO(n + enc + b'junk'))
>>> s == t
True
>>> read_unicodestring1(io.BytesIO(n + enc[:-1]))
Traceback (most recent call last):
...
ValueError: expected 7 bytes in a unicodestring1, but only 6 remain
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surrogatepassz9expected %d bytes in a unicodestring1, but only %d remainN)rrrrGr)rrrrrr�read_unicodestring1osrK�unicodestring1aAA counted Unicode string.
The first argument is a 1-byte little-endian signed int
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cCst|�}|tjkr.td|��n|j|�}t|�|kr_t|dd�Std|t|�f��dS)a�
>>> import io
>>> s = 'abcd\uabcd'
>>> enc = s.encode('utf-8')
>>> enc
b'abcd\xea\xaf\x8d'
>>> n = bytes([len(enc), 0, 0, 0]) # little-endian 4-byte length
>>> t = read_unicodestring4(io.BytesIO(n + enc + b'junk'))
>>> s == t
True
>>> read_unicodestring4(io.BytesIO(n + enc[:-1]))
Traceback (most recent call last):
...
ValueError: expected 7 bytes in a unicodestring4, but only 6 remain
z+unicodestring4 byte count > sys.maxsize: %dzutf-8rJz9expected %d bytes in a unicodestring4, but only %d remainN)r$rArBrrrrG)rrrrrr�read_unicodestring4�srM�unicodestring4aAA counted Unicode string.
The first argument is a 4-byte little-endian signed int
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cCst|�}|tjkr.td|��n|j|�}t|�|kr_t|dd�Std|t|�f��dS)a�
>>> import io
>>> s = 'abcd\uabcd'
>>> enc = s.encode('utf-8')
>>> enc
b'abcd\xea\xaf\x8d'
>>> n = bytes([len(enc)]) + bytes(7) # little-endian 8-byte length
>>> t = read_unicodestring8(io.BytesIO(n + enc + b'junk'))
>>> s == t
True
>>> read_unicodestring8(io.BytesIO(n + enc[:-1]))
Traceback (most recent call last):
...
ValueError: expected 7 bytes in a unicodestring8, but only 6 remain
z+unicodestring8 byte count > sys.maxsize: %dzutf-8rJz9expected %d bytes in a unicodestring8, but only %d remainN)r'rArBrrrrG)rrrrrr�read_unicodestring8�srO�unicodestring8aAA counted Unicode string.
The first argument is a 8-byte little-endian signed int
giving the number of bytes in the string, and the second
argument-- the UTF-8 encoding of the Unicode string --
contains that many bytes.
cCsBt|dddd�}|dkr(dS|dkr8dSt|�S)z�
>>> import io
>>> read_decimalnl_short(io.BytesIO(b"1234\n56"))
1234
>>> read_decimalnl_short(io.BytesIO(b"1234L\n56"))
Traceback (most recent call last):
...
ValueError: invalid literal for int() with base 10: b'1234L'
r2Fr3s00s01T)r5�int)r�srrr�read_decimalnl_short�srScCsKt|dddd�}|dd�dkrA|dd�}nt|�S) z�
>>> import io
>>> read_decimalnl_long(io.BytesIO(b"1234L\n56"))
1234
>>> read_decimalnl_long(io.BytesIO(b"123456789012345678901234L\n6"))
123456789012345678901234
r2Fr3rN�Lr,r,)r5rQ)rrRrrr�read_decimalnl_long�srU�decimalnl_shorta�A newline-terminated decimal integer literal.
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is read.
�decimalnl_longz�A newline-terminated decimal integer literal.
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>>> import io
>>> read_floatnl(io.BytesIO(b"-1.25\n6"))
-1.25
r2Fr3)r5�float)rrRrrr�read_floatnl'srY�floatnla�A newline-terminated decimal floating literal.
In general this requires 17 significant digits for roundtrip
identity, and pickling then unpickling infinities, NaNs, and
minus zero doesn't work across boxes, or on some boxes even
on itself (e.g., Windows can't read the strings it produces
for infinities or NaNs).
cCsB|jd�}t|�dkr2td|�dStd��dS)z�
>>> import io, struct
>>> raw = struct.pack(">d", -1.25)
>>> raw
b'\xbf\xf4\x00\x00\x00\x00\x00\x00'
>>> read_float8(io.BytesIO(raw + b"\n"))
-1.25
r&z>drz(not enough data in stream to read float8N)rrrr)rrrrr�read_float8=s
r[�float8aAn 8-byte binary representation of a float, big-endian.
The format is unique to Python, and shared with the struct
module (format string '>d') "in theory" (the struct and pickle
implementations don't share the code -- they should). It's
strongly related to the IEEE-754 double format, and, in normal
cases, is in fact identical to the big-endian 754 double format.
On other boxes the dynamic range is limited to that of a 754
double, and "add a half and chop" rounding is used to reduce
the precision to 53 bits. However, even on a 754 box,
infinities, NaNs, and minus zero may not be handled correctly
(may not survive roundtrip pickling intact).
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>>> import io
>>> read_long1(io.BytesIO(b"\x00"))
0
>>> read_long1(io.BytesIO(b"\x02\xff\x00"))
255
>>> read_long1(io.BytesIO(b"\x02\xff\x7f"))
32767
>>> read_long1(io.BytesIO(b"\x02\x00\xff"))
-256
>>> read_long1(io.BytesIO(b"\x02\x00\x80"))
-32768
z'not enough data in stream to read long1)rrrrr])rrrrrr�
read_long1cs
r^�long1aA binary long, little-endian, using 1-byte size.
This first reads one byte as an unsigned size, then reads that
many bytes and interprets them as a little-endian 2's-complement long.
If the size is 0, that's taken as a shortcut for the long 0L.
cCset|�}|dkr+td|��n|j|�}t|�|kr[td��nt|�S)ag
>>> import io
>>> read_long4(io.BytesIO(b"\x02\x00\x00\x00\xff\x00"))
255
>>> read_long4(io.BytesIO(b"\x02\x00\x00\x00\xff\x7f"))
32767
>>> read_long4(io.BytesIO(b"\x02\x00\x00\x00\x00\xff"))
-256
>>> read_long4(io.BytesIO(b"\x02\x00\x00\x00\x00\x80"))
-32768
>>> read_long1(io.BytesIO(b"\x00\x00\x00\x00"))
0
rzlong4 byte count < 0: %dz'not enough data in stream to read long4)r"rrrr])rrrrrr�
read_long4�sr`�long4a�A binary representation of a long, little-endian.
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If not isinstance(callable, type), REDUCE complains unless the
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In addition, all the objects on the stack following the topmost
markobject are gathered into a tuple and popped (along with the
topmost markobject), just as for the TUPLE opcode.
Now it gets complicated. If all of these are true:
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at the start).
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then we want to create an old-style class instance without invoking
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Optional arg 'memo' is a Python dict, used as the pickle's memo. It
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Passing the same memo object to another dis() call then allows disassembly
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Optional arg 'indentlevel' is the number of blanks by which to indent
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The value given to 'annotate' must be an integer and is used as a
hint for the column where annotation should start. The default
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In addition to printing the disassembly, some sanity checks are made:
+ All embedded opcode arguments "make sense".
+ Explicit and implicit pop operations have enough items on the stack.
+ When an opcode implicitly refers to a markobject, a markobject is
actually on the stack.
+ A memo entry isn't referenced before it's defined.
+ The markobject isn't stored in the memo.
+ A memo entry isn't redefined.
Nr� z%5d:�end�filez %-4s %s%sr�z(MARK at unknown opcode offset)z(MARK at %d)rzno MARK exists on stackr�r�r�r�zmemo key %r already definedz'stack is empty -- can't store into memoz"can't store markobject in the memor�r�r�z&memo key %r has never been stored into�
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markobject�pop�indexrr�splitr��extend)r�r��memo�indentlevel�annotate�stackZmaxprotoZ markstackZindentchunkZerrormsgZannocolr�rsr��lineZbefore�afterZnumtopopZmarkmsgZmarkposZmemo_idxrrrr) s�-
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>>> import pickle
>>> x = [1, 2, (3, 4), {b'abc': "def"}]
>>> pkl0 = pickle.dumps(x, 0)
>>> dis(pkl0)
0: ( MARK
1: l LIST (MARK at 0)
2: p PUT 0
5: L LONG 1
9: a APPEND
10: L LONG 2
14: a APPEND
15: ( MARK
16: L LONG 3
20: L LONG 4
24: t TUPLE (MARK at 15)
25: p PUT 1
28: a APPEND
29: ( MARK
30: d DICT (MARK at 29)
31: p PUT 2
34: c GLOBAL '_codecs encode'
50: p PUT 3
53: ( MARK
54: V UNICODE 'abc'
59: p PUT 4
62: V UNICODE 'latin1'
70: p PUT 5
73: t TUPLE (MARK at 53)
74: p PUT 6
77: R REDUCE
78: p PUT 7
81: V UNICODE 'def'
86: p PUT 8
89: s SETITEM
90: a APPEND
91: . STOP
highest protocol among opcodes = 0
Try again with a "binary" pickle.
>>> pkl1 = pickle.dumps(x, 1)
>>> dis(pkl1)
0: ] EMPTY_LIST
1: q BINPUT 0
3: ( MARK
4: K BININT1 1
6: K BININT1 2
8: ( MARK
9: K BININT1 3
11: K BININT1 4
13: t TUPLE (MARK at 8)
14: q BINPUT 1
16: } EMPTY_DICT
17: q BINPUT 2
19: c GLOBAL '_codecs encode'
35: q BINPUT 3
37: ( MARK
38: X BINUNICODE 'abc'
46: q BINPUT 4
48: X BINUNICODE 'latin1'
59: q BINPUT 5
61: t TUPLE (MARK at 37)
62: q BINPUT 6
64: R REDUCE
65: q BINPUT 7
67: X BINUNICODE 'def'
75: q BINPUT 8
77: s SETITEM
78: e APPENDS (MARK at 3)
79: . STOP
highest protocol among opcodes = 1
Exercise the INST/OBJ/BUILD family.
>>> import pickletools
>>> dis(pickle.dumps(pickletools.dis, 0))
0: c GLOBAL 'pickletools dis'
17: p PUT 0
20: . STOP
highest protocol among opcodes = 0
>>> from pickletools import _Example
>>> x = [_Example(42)] * 2
>>> dis(pickle.dumps(x, 0))
0: ( MARK
1: l LIST (MARK at 0)
2: p PUT 0
5: c GLOBAL 'copy_reg _reconstructor'
30: p PUT 1
33: ( MARK
34: c GLOBAL 'pickletools _Example'
56: p PUT 2
59: c GLOBAL '__builtin__ object'
79: p PUT 3
82: N NONE
83: t TUPLE (MARK at 33)
84: p PUT 4
87: R REDUCE
88: p PUT 5
91: ( MARK
92: d DICT (MARK at 91)
93: p PUT 6
96: V UNICODE 'value'
103: p PUT 7
106: L LONG 42
111: s SETITEM
112: b BUILD
113: a APPEND
114: g GET 5
117: a APPEND
118: . STOP
highest protocol among opcodes = 0
>>> dis(pickle.dumps(x, 1))
0: ] EMPTY_LIST
1: q BINPUT 0
3: ( MARK
4: c GLOBAL 'copy_reg _reconstructor'
29: q BINPUT 1
31: ( MARK
32: c GLOBAL 'pickletools _Example'
54: q BINPUT 2
56: c GLOBAL '__builtin__ object'
76: q BINPUT 3
78: N NONE
79: t TUPLE (MARK at 31)
80: q BINPUT 4
82: R REDUCE
83: q BINPUT 5
85: } EMPTY_DICT
86: q BINPUT 6
88: X BINUNICODE 'value'
98: q BINPUT 7
100: K BININT1 42
102: s SETITEM
103: b BUILD
104: h BINGET 5
106: e APPENDS (MARK at 3)
107: . STOP
highest protocol among opcodes = 1
Try "the canonical" recursive-object test.
>>> L = []
>>> T = L,
>>> L.append(T)
>>> L[0] is T
True
>>> T[0] is L
True
>>> L[0][0] is L
True
>>> T[0][0] is T
True
>>> dis(pickle.dumps(L, 0))
0: ( MARK
1: l LIST (MARK at 0)
2: p PUT 0
5: ( MARK
6: g GET 0
9: t TUPLE (MARK at 5)
10: p PUT 1
13: a APPEND
14: . STOP
highest protocol among opcodes = 0
>>> dis(pickle.dumps(L, 1))
0: ] EMPTY_LIST
1: q BINPUT 0
3: ( MARK
4: h BINGET 0
6: t TUPLE (MARK at 3)
7: q BINPUT 1
9: a APPEND
10: . STOP
highest protocol among opcodes = 1
Note that, in the protocol 0 pickle of the recursive tuple, the disassembler
has to emulate the stack in order to realize that the POP opcode at 16 gets
rid of the MARK at 0.
>>> dis(pickle.dumps(T, 0))
0: ( MARK
1: ( MARK
2: l LIST (MARK at 1)
3: p PUT 0
6: ( MARK
7: g GET 0
10: t TUPLE (MARK at 6)
11: p PUT 1
14: a APPEND
15: 0 POP
16: 0 POP (MARK at 0)
17: g GET 1
20: . STOP
highest protocol among opcodes = 0
>>> dis(pickle.dumps(T, 1))
0: ( MARK
1: ] EMPTY_LIST
2: q BINPUT 0
4: ( MARK
5: h BINGET 0
7: t TUPLE (MARK at 4)
8: q BINPUT 1
10: a APPEND
11: 1 POP_MARK (MARK at 0)
12: h BINGET 1
14: . STOP
highest protocol among opcodes = 1
Try protocol 2.
>>> dis(pickle.dumps(L, 2))
0: \x80 PROTO 2
2: ] EMPTY_LIST
3: q BINPUT 0
5: h BINGET 0
7: \x85 TUPLE1
8: q BINPUT 1
10: a APPEND
11: . STOP
highest protocol among opcodes = 2
>>> dis(pickle.dumps(T, 2))
0: \x80 PROTO 2
2: ] EMPTY_LIST
3: q BINPUT 0
5: h BINGET 0
7: \x85 TUPLE1
8: q BINPUT 1
10: a APPEND
11: 0 POP
12: h BINGET 1
14: . STOP
highest protocol among opcodes = 2
Try protocol 3 with annotations:
>>> dis(pickle.dumps(T, 3), annotate=1)
0: \x80 PROTO 3 Protocol version indicator.
2: ] EMPTY_LIST Push an empty list.
3: q BINPUT 0 Store the stack top into the memo. The stack is not popped.
5: h BINGET 0 Read an object from the memo and push it on the stack.
7: \x85 TUPLE1 Build a one-tuple out of the topmost item on the stack.
8: q BINPUT 1 Store the stack top into the memo. The stack is not popped.
10: a APPEND Append an object to a list.
11: 0 POP Discard the top stack item, shrinking the stack by one item.
12: h BINGET 1 Read an object from the memo and push it on the stack.
14: . STOP Stop the unpickling machine.
highest protocol among opcodes = 2
a=
>>> import pickle
>>> import io
>>> f = io.BytesIO()
>>> p = pickle.Pickler(f, 2)
>>> x = [1, 2, 3]
>>> p.dump(x)
>>> p.dump(x)
>>> f.seek(0)
0
>>> memo = {}
>>> dis(f, memo=memo)
0: \x80 PROTO 2
2: ] EMPTY_LIST
3: q BINPUT 0
5: ( MARK
6: K BININT1 1
8: K BININT1 2
10: K BININT1 3
12: e APPENDS (MARK at 5)
13: . STOP
highest protocol among opcodes = 2
>>> dis(f, memo=memo)
14: \x80 PROTO 2
16: h BINGET 0
18: . STOP
highest protocol among opcodes = 2
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