Creating a buffer class by extending Python str

Question

I guess this is best explained by code and comments:

import struct

class binary_buffer(str):
    def __init__(self, msg=""):
        self = msg
    def write_ubyte(self, ubyte):
        self += struct.pack("=B", ubyte)
        return len(self)

Output
>> bb = binary_buffer()
>> bb # Buffer starts out empty, as it should
''
>> bb.write_ubyte(200)
1   # We can see that we've successfully written one byte to the buffer
>> bb
''  # Huh? We just wrote something, but where has it gone?

phihag · Accepted Answer · 2013-02-02 09:13:02Z

4

strs are immutable. Therefore,

self += struct.pack("=B", ubyte)

is evaluated as

self = self + struct.pack("=B", ubyte)

This assigns a new value to the name self, but self is just a name like any other. As soon as the method exits, the name (and the associated object) are forgotten.

You want a bytearray:

>>> bb = bytearray()
>>> bb.append(200)
>>> bb
bytearray(b'\xc8')

answered Feb 2, 2013 at 9:13

phihag

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5 Comments

Ulrich Eckhardt Over a year ago

I'd say that in general, assigning to "self" is wrong. At least I can't come up with a legitimate use for this.

Peter Over a year ago

@doomster: Why is that? And how else would you do it?

phihag Over a year ago

@TahitiPetey Because you're just binding something to the name self, which is usually your current object. To change the current object, assign to one of its properties, or, equivalently, call one of its methods.

Peter Over a year ago

@phihag: So when I simply use bb, what method is it invoking to return the bytearray? Is not using self? I would like to be able to pass the binary_buffer object into a socket.sendto() method as the message. Is there a more 'pythonic' way of doing this?

phihag Over a year ago

@TahitiPetey bytearray.append(x) is (approximately) implemented as self._bytes += x. You can pass in a bytearray object to sendto, but you can always create an immutable bytes object from it with bytes(bb).

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