Well, Part 1 of this blog explained about the basics of base 64, here I would like to discuss more about base 64.
Base 64 Padding
I’m starting from example rather explaining in words. In Part 1 I took 3 letter word ‘man’ to explain, now I am taking a 4 letter word ‘many’. Lets do it again,
- The binary representation of each letter is
m 01101101 a 01100001 n 01101110 y 01111001 - Now, group 6-bits, you will get five groups of 6-bits and two bits remaining. See the following table.
011011 010110 000101 101110 011110 01 - Now, how do you get base 64 alphabet for the last 2-bit group? This is where the padding helps. To get exactly 6-bits in each group, keep increasing the binary string (filled with zeroes) to its right hand side. See the following table after increasing 1 byte The newly added bits are in green color.
011011 010110 000101 101110 011110 010000 0000 - Here also, the last group is not 6-bits, so add one more byte, resulting in the following table.
011011 010110 000101 101110 011110 010000 000000 000000 Now, we are exactly getting 6-bits in each group.
This is why padding is needed in base 64 whenever the base 64 string is not divisible exactly groups of 6-bits each. Well, the next step is find the base 64 alphabet for each group. Take the equivalent decimal value of each group.
| bit groups | 011011 | 010110 | 000101 | 101110 | 011110 | 010000 | 000000 | 000000 |
| equivalent decimal (ie. index) | 27 | 22 | 5 | 46 | 30 | 16 | [pad] | [pad] |
|---|---|---|---|---|---|---|---|---|
| encoded value | b | W | F | u | e | Q | = | = |