So, for example I have a string which binary value is exactly 512 - 64 bits long, so the binary conversion of the string does not have to be padded. So that means I have a string of 56 chars.
The binary equivalant of 56 is
00111000, which is 64 - 8 short of 64 bits. How are the remaining bits filled? Are they padded as well?
This is what happens when padding the string:
The message is "padded" (extended) so that its length (in bits) is congruent to 448, modulo 512. That is, the message is extended so that it is just 64 bits shy of being a multiple of 512 bits long. Padding is always performed, even if the length of the message is already congruent to 448, modulo 512.
Padding is performed as follows: a single "1" bit is appended to the message, and then "0" bits are appended so that the length in bits of the padded message becomes congruent to 448, modulo 512. In all, at least one bit and at most 512 bits are appended.
However, in the same document, no clarification is given on the remaining 'empty' bits (512 - 56 * 8 - 8):
A 64-bit representation of b (the length of the message before the padding bits were added) is appended to the result of the previous step. In the unlikely event that b is greater than 2^64, then only the low-order 64 bits of b are used. (These bits are appended as two 32-bit words and appended low-order word first in accordance with the previous conventions.)
At this point the resulting message (after padding with bits and with b) has a length that is an exact multiple of 512 bits. Equivalently, this message has a length that is an exact multiple of 16 (32-bit) words. Let M[0 ... N-1] denote the words of the resulting message, where N is a multiple of 16.
I hope my question is understood?
My guess is that 64 - bits(
00111000) 0's are padded in front of the 64bit representation of the length.