bcrypt: simplify bcrypt's base64 by using base64.Nopadding

[kurenaif]

In the existing bcrypt, pad and unpad by the base64Encode and base64Decode methods of bcrypt/base64. However, in the current encoding/base64 implementation, WithPadding(base64) can omit them.

Also, in the current implementation, if src is 0 bytes at Encode, dst[n-1] == '=' will result in an out-of-range. current base64Encode is not affected by this because it handle either fixed-length salt(16bytes) or hash values(23bytes).

This pull request simplifies the code and reduce the risk of future out-of-range.

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[ufukty]

I was testing the bcrypt/base64 and discovered the issue with paddings.

If anyone there is not convinced yet by the existence of problem. Take a look to spec of base64: https://datatracker.ietf.org/doc/html/rfc4648#section-9

Current code produces "SGVsbG8gd29ybGQu" for "Hello world." which is correct. But when decoding; it adds 3 equal signs (=) at the end of array. It assumes when the length of input is multiples of 4, it should add (=)s.

https://github.com/golang/crypto/blob/630584e8d5aaa1472863b49679b2d5548d80dcba/bcrypt/base64.go#L24

[ufukty]

Correct operation should calculate number of equals as (4 - len(src)) % 4

number of bytes in the decoded string -> number of equals to add
0 -> 0
1 -> 3
2 -> 2
3 -> 1
4 -> 0 (repeating from top)
5 -> 3
6 -> 2
7 -> 1
8 -> 0

Edit: I realized % operator doesn't work as I thought, it produces negative values.

Edit 2: x = len(src); Subtracting x from the smallest number greater than x that is multiple of 4 seems work (x - x % 4 + 4 - x) % 4. I'm aware it looks ridiculous.