32位架构（GCC）上的8位int与32位int

Question

While coding, I try not to use more variable memory than needed and that leads me to write code like this:

for (uint8 i = 0; i < 32; i++) {
   ...
}

instead of:

for (int i = 0; i < 32; i++) {
   ...
}

( uint8 instead of int because i only need to go up to 32)

This would make sense when coding on an 8bit microprocessor. However, am I saving any resources if running this code on a 32bit microprocessor (where int might be 16 or 32 bit)? Does a compiler like GCC do any magic/juju underneath when I explicitly use an 8bit int on a 32bit architecture?

Answer 1

In most cases, there won't be any memory usage difference because i will never be in memory. i will be stored in a CPU register, and you can't really use one register to store two variables. So i will take one register, uint8 or uint32 doesn't matter.

In some rare cases, i will actually be stored in memory, because the loop is so big that all the CPU registers are taken. In this case, there is still a good chance you won't gain any memory either, because others multi-bytes variables will be aligned, and i will be followed by some useless padding bytes to align the next variable.

Now, if i is actually stored in memory, and there are other 8-bit variables to fill the padding, you may save some memory, but it's so little and so unlikely that it probably isn't worth it. Performance-wise, the difference between 8-bit and 32-bit is very architecture dependent, but usually it will be the same.

Also, if you are working in a 64-bit environment, this memory saving is probably non-existent, because the 64-bit call convention impose a huge 16-bytes alignment on the stack (where i will be stored if it is in memory).