Python - What is the most efficient way to generate padding?

Question

Here's the problem: I'm reading binary files in fairly large blocks (512 KiB) and wish to pad the last block with zeros whenever it is shorter than the block size.

Currently, I'm doing something like this:

bytes = f.read(self.chunksize)
if len(bytes) > 0:
    len_diff = self.chunksize - len(bytes)
    if len_diff > 0:
        bytes += reduce(lambda x,y: x+y, ["\0" for i in range(0, len_diff)])

Obviously this is terribly inefficient since that reduce will make a lot of string concatenations. I'm wondering though, how can I achieve this with Python? In C, I'd simply calloc and be done with it.

If it isn't achievable with Python, I'm willing to convert this code into a C module and/or abandon Python entirely for this project, since it's still on the early stages.

Cheers!

EDIT : I'm feeling terrible for not remembering to use the * operator. :-)

This solution worked perfectly for me:

bytes += "\0" * len_diff

EDIT #2 : Using ljust() instead simplified my code a bit, so the correct answer goes to Jeff.

Answer 1

Couldn't you just use ljust() to do the padding since we're dealing with string objects here?

bytes = f.read(self.chunksize)
if bytes:
    bytes = bytes.ljust(self.chunksize, '\0')

Answer 2

bytes += "\0"*len_diff 

应该有帮助

Answer 3

try this.

bytes = "\0" * self.chunksize
rbytes = f.read(self.chunksize)
bytes[:len(rbytes)] = rbytes

or

bytes = f.read(self.chunksize)
bytes += "\0" * (self.chunksize - len(bytes))

Answer 4

怎么样：

bytes += "\0"*len_diff

Answer 5

I needed to pad something encrypted. Here's how to do it.

from Crypto.Util.Padding import pad

...
_bytes = pad(_bytes, block_size)