Did you ever saw a char and thought: "Damn, 1 byte for a single char is pretty darn inefficient"? No? Well I did. So what I decided to do instead is to pack 5 chars, convert each char to a 2 digit integer and then concat those 5 2 digit ints together into one big unsigned int and boom, I saved 5 chars using only 4 instead of 5 bytes. The reason this works is, because one unsigned int is a ten digit long number and so I can save one char using 2 digits. In theory you could save 32 different chars using this technique (the first two digits of an unsigned int are 42 and if you dont want to account for a possible 0 in the beginning you end up with 32 chars). If you would decide to use all 10 digits you could save exactly 3 chars. Why should anyone do that? Idk. Is it way to much work to be useful? Yes. Was it funny? Yes.
Anyone whos interested in the code: Heres how I did it in C: https://pastebin.com/hDeHijX6
Yes I know, the code is probably bad, but I do not care. It was just a funny useless idea I had.
Well it’s certainly possible to fit both uppercase and lowercase + 11 additional characters inside an int (26 + 26 + 11 = 63). The you need a null terminating char, which adds up to 64, which is 6 bits.
So all you need is 6 bits per char. 6 * 5 = 30, which is less than 32.
It’s easier to do this by thinking in binary rather than decimals. Look into bit shifting and other bitwise operations.
Depending on the use-case you might also want to add special case value like @Redkey@programming.dev did in their example, and get kind of UTF-8 pages. Then you can pack lowercase to 5 bits, and uppercase and some special symbols to 10 bits, and it will be smaller if uppercase are rare