Is this how the + operator is implemented in C?

Question

When understanding how primitive operators such as +, -, * and / are implemented in C, I found the following snippet from an

Answer 1

Just out of interest, on the Atmega328P processor, with the avr-g++ compiler, the following code implements adding one by subtracting -1 :

volatile char x;
int main ()
  {
  x = x + 1;  
  }

Generated code:

00000090 
:
volatile char x;
int main ()
  {
  x = x + 1;  
  90:   80 91 00 01     lds r24, 0x0100
  94:   8f 5f           subi    r24, 0xFF   ; 255
  96:   80 93 00 01     sts 0x0100, r24
  }
  9a:   80 e0           ldi r24, 0x00   ; 0
  9c:   90 e0           ldi r25, 0x00   ; 0
  9e:   08 95           ret

Notice in particular that the add is done by the subi instruction (subtract constant from register) where 0xFF is effectively -1 in this case.

Also of interest is that this particular processor does not have a addi instruction, which implies that the designers thought that doing a subtract of the complement would be adequately handled by the compiler-writers.

Does this take advantage of two's complement or other implementation-dependent features?

It would probably be fair to say that compiler-writers would attempt to implement the wanted effect (adding one number to another) in the most efficient way possible for that particularly architecture. If that requires subtracting the complement, so be it.