3 ms·

> Let’s begin by the base64 notation. The cardinality of each byte is 64 so it takes 3 bytes in base64 to represent 2 bytes of actual value. Wait, what? I thou

• rawling 5y ago
  Maybe they're using a 9-bit byte? :)
• brabel 5y ago
  base64 means the "vocabulary" used has 6 bits (2^6 = 64). Hence, to complete a full number of bytes (without using any padding), you need 4 b64-letters: 4 * 6 = 24 = 8 * 3 So, you're correct... the exact amount of bytes that it takes to represent "actual" bytes goes like this: Actual bytes | b64 bytes required | overhead 1 | 2 | 2x 2 | 3 | 1.5x 3 | 4 | 1.33x 4 | 6 | 1.5x 5 | 7 | 1.4x 6 | 8 | 1.33x 7 | 10 | 1.43x 8 | 11 | 1.37x 9 | 12 | 1.33x EDIT: As you can see, this averages with an overhead of between 33% (best case scenario where the encoding requires no padding, happens every 3 rows above) and something like 37%, decreasing with the number of bytes being encoded and approaching the minimum, 33% (e.g. to encode 1024 bytes, you need 1366 b64 digits, an overhead of 1.333984375x).