a + = b和a = a + b之间的性能差异

Question

I know the two statements below give the same results:

a+=b; 
a=a+b;

In the case of a += b , a is evaluated only once while in the case of a = a + b , a is evaluated twice.

Is there any difference in performance between the two? If not, are there any variations of the above where there is a difference?

Answer 1

From the standard (section 6.5.16.2, point 3):

A compound assignment of the form E1 op = E2 differs from the simple assignment expression *E1 = E1 op (E2) only in that the lvalue E1 is evaluated only once.

That is, if instead you are doing

*a += 1

it will only determine the target location once instead of twice. For "simple" variables as in your example, it might not make much of a difference. Of course, if the compiler knows there is no need to do it twice, it can still optimise and do it once. In case there are other entities that can change the pointer (such as another thread), there is a real difference.

EDIT: Perhaps a better example is something weird like the following (abusing the fact that &b == &a-1 in my case):

int a, b, *p;

a = 1; b = 5; p = &a;
*p += (--p == &b) ? 1 : 0;
printf("%d %d\n",a,b); // prints 1 6, because --p happens first, then *p

a = 1; b = 5; p = &a;
*p = *p + ((--p == &b) ? 1 : 0);
printf("%d %d\n",a,b); // prints 1 2, because the second *p is evaluated first,
                       // then --p and then the first *p

Answer 2

a+=1 is probably the best way to go (However the difference in performance is almost insignificant). You should look up for other parts of your code if you are trying to achieve some performance improvements.