In this function, what is the real meaning of the " * " operator?

Question

from Crypto.Cipher import AES
import os
import base64
import socket
import subprocess

BLOCK_SIZE = 16
PADDING='#'

def _pad(data, pad_with=PADDING):
    return data + (BLOCK_SIZE - len(data) % BLOCK_SIZE) * PADDING

I understand that the character "#" is used to pad the block but I can't understand the meaning of the "*", used normally for multiplication.

Answer 1

In Python, you can multiply a string by an integer to repeat the string that many times. E.g.

'a' * 4

produces

'aaaa'

Any class can overload an operator like * by implementing specially-named methods.

Answer 2

str type gets the * operator:

'#' * 10 = '##########'

because __mul__() method is defined in the str class definition:

'#'.__mul__(10) = '##########'

Answer 3

The * PADDING means exactly that (times PADDING).

* PADDING is add (BLOCK_SIZE - len(data) % BLOCK_SIZE) times # to the end of data.
It calculates the amount of padding needed for the specific data based on its size and the BLOCK_SIZE and adds extra # which is the PADDING.

The calculation is the BLOCK_SIZE minus the length of the data passed to the function, all this modulo the BLOCK_SIZE.

Example:

>>> BLOCK_SIZE = 20
>>> data = 'this is my data'
>>> PADDING = '#'
>>> data + (BLOCK_SIZE - len(data) % BLOCK_SIZE) * PADDING
'this is my data#####'

The calculation performed is:

data + (20 - 15%20)*PADDING = data + (20 - 15)*PADDING = data + 5*PADDING = data + '#####'

Answer 4

>>> a= '#'
>>> a* 12
'############'
>>>