int a = 2;
printf("%d", ++a + ++a + ++a);

> 13

printf("%d", ++a + (++a + ++a));

CS>int a = 2;
CS>printf("%d", ++a + ++a + ++a);
CS>

>> 13
CS>

    +
   / \
  /   \
++a(1) ++a(2)
        \
         \
          +
           \
            \
          ++a(3)

inc a           ; a = 3 (1) в скобочках - порядковый номер инкремента в примере
inc a           ; a = 4 (2)
mov eax, a      ; a = 4, eax = 4 
add eax, a      ; a = 4, eax = 8
inc a           ; a = 5 (3), eax = 8
add eax, a      ; a = 5, eax = 13

csm1.c:5:24: warning: multiple unsequenced modifications to 'a' [-Wunsequenced]
        printf("%d\n", ++a + ++a + ++a);
                       ^     ~~

extern int a,b;
void f(void) {
  a = (a++) % b;
}

ISO/IEC/14882-2011
1.9/15
Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced. [ Note: In an expression that is evaluated more than once during the execution of a program, unsequenced and indeterminately sequenced evaluations of its subexpressions need not be performed consistently in different evaluations. —end note ] The value computations of the operands of an operator are sequenced before the value computation of the result of the operator. If a side effect on a scalar object is unsequenced relative to either anotherside effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined. [ Example:

void f(int, int);
void g(int i, int* v) {
i = v[i++]; // the behavior is undefined
i = 7, i++, i++; // i becomes 9
i = i++ + 1; // the behavior is undefined
i = i + 1; // the value of i is incremented
f(i = -1, i = -1); // the behavior is undefined
}