Most code I have ever read uses a int
for standard error handling (return values from functions and such). But I am wondering if there is any benefit to be had from using a uint_8
will a compiler -- read: most C compilers on most architectures -- produce instructions using the immediate address mode -- i.e., embed the 1-byte integer into the instruction ? The key instruction I'm thinking about is the compare after a function, using uint_8 as its return type, returns.
I could be thinking about things incorrectly, as introducing a 1 byte type just causes alignment issues -- there is probably a perfectly sane reason why compiles like to pack things in 4-bytes and this is possibly the reason everyone just uses ints -- and since this is stack related issue rather than the heap there is no real overhead.
Doing the right thing is what I'm thinking about. But lets say say for the sake of argument this is a popular cheap microprocessor for a intelligent watch and that it is configured with 1k of memory but does have different addressing modes in its instruction set :D
Another question to slightly specialize the discussion (x86) would be: is the literal in:
uint_32 x=func(); x==1;
and
uint_8 x=func(); x==1;
the same type ? or will the compiler generate a 8-byte literal in the second case. If so it may use it to generate a compare instruction which has the literal as an immediate value and the returned int as a register reference. See CMP instruction types..