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Edit: pickletools.cpython-39.pyc

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q�dS�(���ar��"Executable documentation" for the pickle module.

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.
�����N)�dis�genops�optimize���������������c�������������������@���s���e�Zd�ZdZdd��ZdS�)�ArgumentDescriptor��name�n�reader�docc�����������������C���s`���t�|t�sJ��||�_t�|t�r8|dks<|tttttfv�s<J��||�_	||�_
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    >>> import io
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    65535
    ����z<Hr���z'not enough data in stream to read uint2N�r$����len�_unpackr%���r&���r���r���r����
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    >>> import io
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    True
    ����z<ir���z&not enough data in stream to read int4Nr,���r&���r���r���r����	read_int4����s����	
r2����int4r1���z8Four-byte signed integer, little-endian, 2's complement.c�����������������C���s0���|���d�}t|�dkr$td|�d�S�td��dS�)z�
    >>> import io
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    True
    r1���z<Ir���z'not enough data in stream to read uint4Nr,���r&���r���r���r����
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    >>> import io
    >>> read_uint8(io.BytesIO(b'\xff\x00\x00\x00\x00\x00\x00\x00'))
    255
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    True
    ����z<Qr���z'not enough data in stream to read uint8Nr,���r&���r���r���r����
read_uint8&��s����	
r7����uint8r6���z+Eight-byte unsigned integer, little-endian.Tc�����������������C���s����|�����}|�d�std��|dd��}|rtdD�]8}|�|�r.|�|�sVtd||f���|dd��}�qtq.td|���|r�t�|�d	��d
�}|S�)au��
    >>> 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'
    ����
z-no newline found when trying to read stringnlNr���)����"����'z,strinq quote %r not found at both ends of %rr#���zno string quotes around %rr����ascii)�readline�endswithr%����
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read_stringnl;��s"����

�rE����stringnlz�A newline-terminated string.

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                   c�����������������C���s���t�|�dd�S�)NF)rC���)rE����r'���r���r���r����read_stringnl_noescapet��s����rH����stringnl_noescapea��A newline-terminated string.

This is a str-style string, without embedded escapes,
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                        c�����������������C���s���dt�|��t�|��f�S�)zp
    >>> import io
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    'Queue Empty'
    z%s %s)rH���rG���r���r���r����read_stringnl_noescape_pair���s����rJ����stringnl_noescape_paira���A pair of newline-terminated strings.

These are str-style strings, without embedded
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                             consist solely of printable ASCII characters.
                             The pair is returned as a single string, with
                             a single blank separating the two strings.
                             c�����������������C���sL���t�|��}|dksJ��|��|�}t|�|kr4|�d�S�td|t|�f���dS�)z�
    >>> import io
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    r����latin-1z2expected %d bytes in a string1, but only %d remainN)r)���r$���r-���rB���r%����r'���r
���r(���r���r���r����read_string1���s����

�rN����string1z�A counted string.

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              c�����������������C���sT���t�|��}|dk�rtd|���|��|�}t|�|kr<|�d�S�td|t|�f���dS�)aP��
    >>> 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"))
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    ...
    ValueError: expected 50331648 bytes in a string4, but only 6 remain
    r���zstring4 byte count < 0: %drL���z2expected %d bytes in a string4, but only %d remainN)r2���r%���r$���r-���rB���rM���r���r���r����read_string4���s����

�rP����string4z�A counted string.

The first argument is a 4-byte little-endian signed int giving
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              c�����������������C���sF���t�|��}|dksJ��|��|�}t|�|kr.|S�td|t|�f���dS�)z�
    >>> import io
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    b'abc'
    r���z1expected %d bytes in a bytes1, but only %d remainN)r)���r$���r-���r%���rM���r���r���r����read_bytes1���s����

�rR����bytes1z�A counted bytes string.

The first argument is a 1-byte unsigned int giving the number
              of bytes, and the second argument is that many bytes.
              c�����������������C���s\���t�|��}|dksJ��|tjkr*td|���|��|�}t|�|krD|S�td|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
    r���z#bytes4 byte count > sys.maxsize: %dz1expected %d bytes in a bytes4, but only %d remainN)r4����sys�maxsizer%���r$���r-���rM���r���r���r����read_bytes4���s����

�rV����bytes4z�A counted bytes string.

The first argument is a 4-byte little-endian unsigned int giving
              the number of bytes, and the second argument is that many bytes.
              c�����������������C���s\���t�|��}|dksJ��|tjkr*td|���|��|�}t|�|krD|S�td|t|�f���dS�)a���
    >>> 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
    r���z#bytes8 byte count > sys.maxsize: %dz1expected %d bytes in a bytes8, but only %d remainN)r7���rT���rU���r%���r$���r-���rM���r���r���r����read_bytes8��s����

�rX����bytes8z�A counted bytes string.

The first argument is an 8-byte little-endian unsigned int giving
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              c�����������������C���s`���t�|��}|dksJ��|tjkr*td|���|��|�}t|�|krHt|�S�td|t|�f���dS�)a���
    >>> import io, struct, sys
    >>> read_bytearray8(io.BytesIO(b"\x00\x00\x00\x00\x00\x00\x00\x00abc"))
    bytearray(b'')
    >>> read_bytearray8(io.BytesIO(b"\x03\x00\x00\x00\x00\x00\x00\x00abcdef"))
    bytearray(b'abc')
    >>> bigsize8 = struct.pack("<Q", sys.maxsize//3)
    >>> read_bytearray8(io.BytesIO(bigsize8 + b"abcdef"))  #doctest: +ELLIPSIS
    Traceback (most recent call last):
    ...
    ValueError: expected ... bytes in a bytearray8, but only 6 remain
    r���z'bytearray8 byte count > sys.maxsize: %dz5expected %d bytes in a bytearray8, but only %d remainN)r7���rT���rU���r%���r$���r-����	bytearrayrM���r���r���r����read_bytearray89��s����

�r[����
bytearray8z�A counted bytearray.

The first argument is an 8-byte little-endian unsigned int giving
              the number of bytes, and the second argument is that many bytes.
              c�����������������C���s0���|�����}|�d�std��|dd��}t|d�S�)zm
    >>> import io
    >>> read_unicodestringnl(io.BytesIO(b"abc\\uabcd\njunk")) == 'abc\uabcd'
    True
    r9���z4no newline found when trying to read unicodestringnlNr���zraw-unicode-escape)r=���r>���r%���r���r&���r���r���r����read_unicodestringnl[��s
����
r]����unicodestringnlz�A newline-terminated Unicode string.

This is raw-unicode-escape encoded, so consists of
                      printable ASCII characters, and may contain embedded
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    >>> 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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    >>> import io
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    >>> 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]))
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    >>> 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
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    >>> import io
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>>> read_decimalnl_short(io.BytesIO(b"1234L\n56"))
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    ...
    ValueError: invalid literal for int() with base 10: b'1234L'
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    >>> import io

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    32767
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    32767
    >>> read_long4(io.BytesIO(b"\x02\x00\x00\x00\x00\xff"))
    -256
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If not isinstance(callable, type), REDUCE complains unless the
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where anyobject may have been mutated, as follows:

If the object has a __setstate__ method,

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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:

+ The argtuple is empty (markobject was at the top of the stack
          at the start).

+ The class object does not have a __getinitargs__ attribute.

then we want to create an old-style class instance without invoking
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Else (the argtuple is not empty, it's not an old-style class object,
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Else (the class object does have a __safe_for_unpickling__ attr),
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Unlike INST, OBJ takes no arguments from the opcode stream.  Instead
      the class object is taken off the stack, immediately above the
      topmost markobject:

Stack before: ... markobject classobject stackslice
      Stack after:  ... new_instance_object

As for INST, the remainder of the stack above the markobject is
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opcode, arg, pos

opcode is an OpcodeInfo record, describing the current opcode.

If the opcode has an argument embedded in the pickle, arg is its decoded
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If the pickle has a tell() method, pos was the value of pickle.tell()
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��|�|��q0td||d��|�rhtd|���dS�)aK��Produce a symbolic disassembly of a pickle.

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Optional arg 'out' is a file-like object to which the disassembly is
    printed.  It defaults to sys.stdout.

Optional arg 'memo' is a Python dict, used as the pickle's memo.  It
    may be mutated by dis(), if the pickle contains PUT or BINPUT opcodes.
    Passing the same memo object to another dis() call then allows disassembly
    to proceed across multiple pickles that were all created by the same
    pickler with the same memo.  Ordinarily you don't need to worry about this.

Optional arg 'indentlevel' is the number of blanks by which to indent
    a new MARK level.  It defaults to 4.

Optional arg 'annotate' if nonzero instructs dis() to add short
    description of the opcode on each line of disassembled output.
    The value given to 'annotate' must be an integer and is used as a
    hint for the column where annotation should start.  The default
    value is 0, meaning no annotations.

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
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+ A memo entry isn't referenced before it's defined.

+ The markobject isn't stored in the memo.

+ A memo entry isn't redefined.
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r���c�������������������@���s���e�Zd�Zdd��ZdS�)�_Examplec�����������������C���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: I    INT        1
    8: a    APPEND
    9: I    INT        2
   12: a    APPEND
   13: (    MARK
   14: I        INT        3
   17: I        INT        4
   20: t        TUPLE      (MARK at 13)
   21: p    PUT        1
   24: a    APPEND
   25: (    MARK
   26: d        DICT       (MARK at 25)
   27: p    PUT        2
   30: c    GLOBAL     '_codecs encode'
   46: p    PUT        3
   49: (    MARK
   50: V        UNICODE    'abc'
   55: p        PUT        4
   58: V        UNICODE    'latin1'
   66: p        PUT        5
   69: t        TUPLE      (MARK at 49)
   70: p    PUT        6
   73: R    REDUCE
   74: p    PUT        7
   77: V    UNICODE    'def'
   82: p    PUT        8
   85: s    SETITEM
   86: a    APPEND
   87: .    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: I    INT        42
  110: s    SETITEM
  111: b    BUILD
  112: a    APPEND
  113: g    GET        5
  116: a    APPEND
  117: .    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
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Edit: pickletools.cpython-39.pyc

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