Discussion
What is the output of this C code?
int main()
{
int a = 1, b = 1, d = 1;
printf("%d, %d, %d", ++a + ++a+a++, a++ + ++b, ++d + d++ + a++);
}
Answer: A
Correct answer: 15, 4, 5
++a + ++a + a++ modifies a three times in one expression, and the three printf arguments may be evaluated in any order - the standard does not fix it. The expression modifies a variable more than once, and reads it as well, with no sequence point in between. The C standard makes that undefined behaviour, so the answer listed is what one common compiler happens to produce - another compiler, or another optimisation level, may give something else, and no program should depend on it. The lesson to take away is not the number but the rule: one variable, one modification per expression.
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