Does while(i--) s+= a[i]; contain undefined behavior in C and C++?

Question

Consider simple code:

#include "stdio.h"

#define N 10U

int main() {
    int a[N] = {0};
    unsigned int i = N;
    int s = 0;

    // Fill a

    while(i--)
        s += a[i];

    printf("Sum is %d\n", s);

    return 0;
}

Does while loop contain undefined behavior because of integer underflow? Do compilers have right to assume that while loop condition is always true because of that and end up with endless loop?

What if i is signed int? Doesn't it contain pitfalls related to array access?

Update

I run this and similar code many times and it worked fine. Moreover, it's popular way to iterate over arrays and vectors backwards. I'm asking this question to make sure that this way is OK from point of view of standard.

At glance, it's obviously not infinite. On other hand, sometimes compiler can "optimize" away some conditions and code assuming that code contains no undefined behavior. It can lead to infinite loops and other unwanted consequences. See this.

Answer 1

This code doesn't invoke undefined behavior. The loop will be terminated once i becomes 0.

For unsigned int, there is no integer over/underflow. The effect will be same with i as signed except there will no wrapping in this case.