I can do this on initialization for a struct Foo:
Foo foo = {bunch, of, things, initialized};
but, I can\'t do this:
Foo f
Memory Footprint - Here is an interesting i386 addition.
After much hassle, using optimization and memcpy seems to generate the smallest footprint using i386 with GCC and C99. I am using -O3 here. stdlib seems to have all sorts of fun compiler optimizations at hand, and this example makes use of that (memcpy is actually compiled out here).
Do this by:
Foo foo; //some global variable
void setStructVal (void) {
const Foo FOO_ASSIGN_VAL = { //this goes into .rodata
.bunch = 1,
.of = 2,
.things = 3,
.initialized = 4
};
memcpy((void*) &FOO_ASSIGN_VAL, (void*) foo, sizeof(Foo));
return;
}
Result:
Example:
Say Foo was a 48 field struct of uint8_t values. It is aligned in memory.
(IDEAL) On a 32-bit machine, this COULD be as quick as 12 MOVL instructions of immediates out to foo's address space. For me this is 12*10 == 120bytes of .text in size.
(ACTUAL) However, using the answer by AUTO will likely generate 48 MOVB instructions in .text. For me this is 48*7 == 336bytes of .text!!
(SMALLEST*) Use the memcpy version above. IF alignment is taken care of,
So, for me at least with my i386 code,
- Ideal: 120 bytes
- Direct: 336 bytes
- Smallest: 288 bytes
*Smallest here means 'smallest footprint I know of'. It also executes faster than the above methods (24 instructions vs 48). Of course, the IDEAL version is fastest & smallest, but I still can't figure that out.
-Justin
*Does anyone know how to get implementation of 'IDEAL' above? It is annoying the hell out of me!!