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

Extensive comments about the pickle protocols and pickle-machine opcodes
can be found here.  Some functions meant for external use:

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@seZdZdZdd�ZdS)�ArgumentDescriptor��name�n�reader�doccCs||_||_||_||_dS�Nr)�selfrr
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    65535
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    >>> import io
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    True
    �z<irz&not enough data in stream to read int4Nr$rrrr�	read_int4�s	
r*�int4r)z8Four-byte signed integer, little-endian, 2's complement.cCs0|�d�}t|�dkr$td|�dStd��dS)z�
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    255
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    True
    r)z<Irz'not enough data in stream to read uint4Nr$rrrr�
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    255
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    �z<Qrz'not enough data in stream to read uint8Nr$rrrr�
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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'"))
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    �
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�r=�stringnlz�A newline-terminated string.

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                   cCst|dd�S)NF)r;)r=�rrrr�read_stringnl_noescapetsr@�stringnl_noescapeaA newline-terminated string.

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    z%s %s)r@r?rrr�read_stringnl_noescape_pair�srB�stringnl_noescape_paira�A pair of newline-terminated strings.

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�rF�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: %drDz2expected %d bytes in a string4, but only %d remainN)r*rrr%r:rErrr�read_string4�s



�rH�string4z�A counted string.

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    >>> import io
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    b'abc'
    z1expected %d bytes in a bytes1, but only %d remainN)r!rr%rrErrr�read_bytes1�s	

�rJ�bytes1z�A counted bytes string.

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    >>> import io
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    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%rErrr�read_bytes4�s



�rN�bytes4z�A counted bytes string.

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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
    z#bytes8 byte count > sys.maxsize: %dz1expected %d bytes in a bytes8, but only %d remainN)r/rLrMrrr%rErrr�read_bytes8s


�rP�bytes8z�A counted bytes string.

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    >>> 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
    z'bytearray8 byte count > sys.maxsize: %dz5expected %d bytes in a bytearray8, but only %d remainN)r/rLrMrrr%�	bytearrayrErrr�read_bytearray89s


�rS�
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rV�unicodestringnlz�A newline-terminated Unicode string.

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    >>> import io
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    >>> 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
    �utf-8�
surrogatepassz9expected %d bytes in a unicodestring1, but only %d remainN)r!rr%rUrrErrr�read_unicodestring1us

�rZ�unicodestring1aAA counted Unicode string.

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                    cCsXt|�}|tjkrtd|��|�|�}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: %drXrYz9expected %d bytes in a unicodestring4, but only %d remainN)r,rLrMrrr%rUrErrr�read_unicodestring4�s


�r\�unicodestring4aAA counted Unicode string.

                    The first argument is a 4-byte little-endian signed int
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                    cCsXt|�}|tjkrtd|��|�|�}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)]) + b'\0' * 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: %drXrYz9expected %d bytes in a unicodestring8, but only %d remainN)r/rLrMrrr%rUrErrr�read_unicodestring8�s


�r^�unicodestring8aBA counted Unicode string.

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                    cCs.t|ddd�}|dkrdS|dkr&dSt|�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'
    F�r:r;s00s01T�r=�int�r�srrr�read_decimalnl_short�srecCs2t|ddd�}|dd�dkr*|dd�}t|�S)z�
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    >>> read_decimalnl_long(io.BytesIO(b"1234L\n56"))
    1234

    >>> read_decimalnl_long(io.BytesIO(b"123456789012345678901234L\n6"))
    123456789012345678901234
    Fr`rN�Lrarcrrr�read_decimalnl_longsrg�decimalnl_shorta�A newline-terminated decimal integer literal.

                          This never has a trailing 'L', and the integer fit
                          in a short Python int on the box where the pickle
                          was written -- but there's no guarantee it will fit
                          in a short Python int on the box where the pickle
                          is read.
                          �decimalnl_longz�A newline-terminated decimal integer literal.

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                         of any size.
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    >>> read_floatnl(io.BytesIO(b"-1.25\n6"))
    -1.25
    Fr`)r=�floatrcrrr�read_floatnl-srk�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
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              cCs0|�d�}t|�dkr$td|�dStd��dS)z�
    >>> import io, struct
    >>> raw = struct.pack(">d", -1.25)
    >>> raw
    b'\xbf\xf4\x00\x00\x00\x00\x00\x00'
    >>> read_float8(io.BytesIO(raw + b"\n"))
    -1.25
    r.z>drz(not enough data in stream to read float8Nr$rrrr�read_float8Cs

rm�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).
             )�decode_longcCs.t|�}|�|�}t|�|kr&td��t|�S)a+
    >>> 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)r!rr%rrorErrr�
read_long1is

rp�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.
    cCsBt|�}|dkrtd|��|�|�}t|�|kr:td��t|�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
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d|��dS)aKProduce a symbolic disassembly of a pickle.

    'pickle' is a file-like object, or string, containing a (at least one)
    pickle.  The pickle is disassembled from the current position, through
    the first STOP opcode encountered.

    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
      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%srr�z(MARK at unknown opcode offset)z(MARK at %d)rzno MARK exists on stack)r�r�r�r�r�z(as %d)zmemo key %r already definedz'stack is empty -- can't store into memoz"can't store markobject in the memo)r�r�r�z&memo key %r has never been stored into�
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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
highest protocol among opcodes = 2
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