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.
Good question! I have read a bit more about it and this does indeed heavily depend on the respective compiler implementation. Depending on the compiler, it may prefer default if-else ladders for few cases. For more, dense cases, LUTs may play a larger role. For less dense cases binary search might be chosen.
I inspected the generated assembler code of your (slightly extended) example via https://godbolt.org/
The code I used:
Using x86-64 gcc 15.2 this leads to a couple of
cmpfollowed byjeinstructions, so “compare” and “jump to label if equal” which basically is a typical if-else ladder. I get the same for x64 msvc v19.43.Changing the cases to 1, 2 and 3 instead of 5000, 2000 and 1000 does not change the outcome.
Increasing to 23 different but incrementally increasing cases (cases 1 to 23) does not change the outcome as well for gcc. But here msvc has introduced a performance optimization: it decreased the input value by one to get a range of 0 to 22 and then created a jump table, so a LUT with execution addresses. (I am not going to detail the assembler commands logic here, but you can use the C++ code below and take a look yourself. :) )
So even in this simple example we can already see how different compilers may implement switch cases differently depending on its structure. Even though gcc chose the apparently less efficient solution here, usually one may trust on the compiler choosing the most efficient switch implementation. ;)
As far as I know, we would not even get the chance of similar optimizations if choosing if-else ladders directly instead of a switch-case structure. It would be interesting to put this to a test though and see whether some compilers translate if-else ladders equivalently with the performance benefits that can currently come with switch structures.
The inflated code: