weird thing in C: not zero is not equal to one

Question

int main()
{
    int a = 2147483647; 
    int c = a + a + 2;
    printf("2a + 2 = %d, !(2a + 2) = %d, !c = %d\n", (2 * a + 2), !(a + a + 2), !c);
}

I got this when I run those codes above

2a + 2 = 0, !(2a + 2) = 0, !c = 1

a is Tmax which in binary form is 0111...111, so a + a + 2 is suposed to be 0000...0000. I don't understand why !(2a + 2) is not equal to 1 when (2a + 2) is actually 0, but !c is 1!!! Please help me

@interjay thanks for help! But I don't get the meaning of undefined behavior. Do you mean that a + a + 2 can be a non-zero value? — user6323131, Jun 02 '22 at 17:16
Undefined behaviour means there's no definition for the outcome, so when you ask whether "Do you mean", then there's no correct answer as there's no formal definition of what the operation should do. — Matthew, Jun 02 '22 at 17:19
@phuclv Thank you for you link, I understand what is going on here~~~ — user6323131, Jun 02 '22 at 17:23
Not sure this is a dup. The question is more about the nature of undefined behavior rather than unsigned vs signed wraparound. — dbush, Jun 02 '22 at 19:40

dbush · Accepted Answer · 2022-06-02T17:22:28.913

Assuming an int is 32 bits, the operations you're performing cause signed integer overflow. Doing so triggers undefined behavior. It does not necessarily wraparound.

In fact, this specific case is given as an example of undefined behavior in section 3.4.3p3 of the C standard where the term undefined behavior is defined:

An example of undefined behavior is the behavior on integer overflow

You've encountered one of the stranger ways that undefined behavior can manifest itself, where seemingly identical operations behave differently. Undefined behavior means that no guarantees are made about what your program will do.

You can not imagine how desprate I am when I saw this output came out — user6323131, Jun 02 '22 at 17:19

weird thing in C: not zero is not equal to one

1 Answers1