vector.size() - 2 leads to infinite loop - C++

Question

This question already has answers here: question from:https://stackoverflow.com/questions/65644986/vector-size-2-leads-to-infinite-loop-c

Answer 1

x.size() returns an unsigned value. If this is smaller than 2 then the value will underflow and the result will become very large.

On the other hand since i is a signed type hen it will sooner or later come to the max value of an int (which will still be less than the result of x.size() - 2), and then i++ will lead to arithmetic overflow which is undefined behavior.

Before you attempt this loop you must make sure that x.size() >= 2, and you need to make sure that the counter variable is an unsigned type (like size_t):

if (x.size() >= 2)
{
    for (size_t i = 0; i < x.size() - 2; ++i)
    {
        // ...
    }

}