Unsigned integers in C++ for loops - c++

Question

I have made some research on Stackoverflow about reverse for loops in C++ that use an unsigned integer instead of a signed one. But I still do NOT understand why there is a problem (see Unsigned int reverse iteration with for loops). Why the following code will yield a segmentation fault?
#include <vector>
#include <iostream>
using namespace std;
int main(void)
{
vector<double> x(10);
for (unsigned int i = 9; i >= 0; i--)
{
cout << "i= " << i << endl;
x[i] = 1.0;
}
cout << "x0= " << x[0] << endl;
return 0;
}
I understand that the problem is when the index i will be equal to zero, because there is something like an overflow. But I think an unsigned integer is allowed to take the zero value, isn't it? Now if I replace it with a signed integer, there is absolutely no problem.
Does somebody can explain me the mechanism behind that reverse loop with an unsigned integer?
Thank you very much!

Answer 1

The problem here is that an unsigned integer is never negative.
Therefore, the loop-test:
i >= 0
will always be true. Thus you get an infinite loop.
When it drops below zero, it wraps around to the largest value unsigned value.
Thus, you will also be accessing x[i] out-of-bounds.
This is not a problem for signed integers because it will simply go negative and thus fail i >= 0.
Thus, if you want to use unsigned integers, you can try one of the following possibilities:
for (unsigned int i = 9; i-- != 0; )
and
for (unsigned int i = 9; i != -1; i--)
These two were suggested by GManNickG and AndreyT from the comments.
And here's my original 3 versions:
for (unsigned int i = 9; i != (unsigned)0 - 1; i--)
or
for (unsigned int i = 9; i != ~(unsigned)0; i--)
or
for (unsigned int i = 9; i != UINT_MAX; i--)

Answer 2

The problem is, your loop allows i to be as low as zero and only expects to exit the loop if i is less than 0. Since i is unsigned, it can never be less than 0. It rolls over to 2^32-1. That is greater than the size of your vector and so results in a segfault.

Answer 3

Whatever the value of unsigned int i it is always true that i >= 0 so your for loop never ends.
In other words, if at some point i is 0 and you decrement it, it still stays non-negative, because it contains then a huge number, probably 4294967295 (that is 232-1).

Answer 4

The problem is here:
for (unsigned int i = 9; i >= 0; i--)
You are starting with a value of 9 for an unsigned int and your exit definition is i >= 0 and this will be always true. (unsigned int will never be negative!!!). Because of this your loop will start over (endless loop, because i=0 then -1 goes max uint).

Answer 5

As you said a decrease of an unsigned below zero, which happens right after the last step of the loop, creates an overflow, the number wraps around to its maximum value and thus we end up with an infinite loop.
Does somebody can explain me the mechanism behind that reverse loop with an unsigned integer?
My preferred method for a reverse loop with an index is this:
for (unsigned int i = 9; i > 0; --i) {
cout << "i= " << x[i - 1] << endl;
}
and that is why because it maps most closely to the normal loop equivalent:
for (unsigned int i = 0; i < 9; ++i) {
cout << "i= " << x[i] << endl;
}
If then you need to access the indexed element multiple times and you don't want to continuously write [i - 1], you can add something like this as the first line in the loop:
auto& my_element = my_vector[i - 1];