C++ Test - Does sign bit 1 always mean negative?

Question

I was just taking a C++ test and I got the following question wrong:

Q: What is the output of the following program?

#include <iostream>
#include <stdint.h>

using namespace std;

int main() {
    int a = 0;
    for (int8_t i = 1; i > 0; i <<= 1)
        a++;
    cout << a;
    return 0;
}

There were the following answers to choose from

• 8
• 7
• Undefined Behavior
• Compile Error

The "correct" answer was 7. If there was "Implementation-Defined Behavior" in the answers, I would choose that, so I chose Undefined Behavior which was sort of the closest. I understand that in sign-and-magnitute, 1's complement, and 2's complement the answer will be 7. But doesn't the C++ standard theoretically allow any other number representations? For example, sign and magnitude, but 0 means negative?

Am I correct in that the real correct answer to this question should be Implementation-Defined Behavior, and if not, could you please explain why the answer is 7 regardless of the implementation?

I read the comments to the question and it appears that initially the type of a was char, which apparently had raised a lot of complaints about whether char is signed or not, so the testsetter changed it to int8_t. As a bonus question, is <stdint.h> part of C++? O_O

Answer 1

I would say it is undefined (and not implementation-defined) for a different reason.

From 5.8:3

The value of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are zero-filled. If E1 has an unsigned type, the value of the result is E1 × 2^E2 , reduced modulo one more than the maximum value representable in the result type. Otherwise, if E1 has a signed type and non-negative value, and E1 × 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.