#### Arithmetic overflow using all 32-bit ints in C++?

This should simply be adding 1 to an unsigned int: `prev = nums[nextIndex + 1];` but it gives a warning saying

Arithmetic overflow: Using operator ‘+’ on a 4 byte value and then casting the result to a 8 byte value. Cast the value to the wider type before calling operator ‘+’ to avoid overflow

But I’m not using any 8 byte values, they are all 32-bit integers… Why is it casting stuff into 8 byte values for me?

I’m doing some questions on LeetCode so here is the full code for the question I’m working on, just rotating a vector of ints:

``````    void rotate(std::vector<int>& nums, int k) {
k %= nums.size();
if (k > 0) {
unsigned int offsetCounter = 0;
int prev;
int next = nums;
for (unsigned int i = 0; i < nums.size(); i++) {

int f = k * i;

unsigned int nextIndex = ((unsigned int)((unsigned int)k * ((unsigned int)i + (unsigned int)1)) + (unsigned int)offsetCounter) % (unsigned int)nums.size();

if (nextIndex == offsetCounter) {
offsetCounter++;
prev = nums[nextIndex + 1];
}
else
{
prev = nums[nextIndex];
}
nums[nextIndex] = next;
next = prev;
}
}
}
``````

`nextIndex` also gave the same warning and the only thing that got rid of it was casting everything to an unsigned int. I don’t understand why it says I am using 8 byte values when I am definitely not using 8 byte values. I’ve ignored warnings like this in the past but LeetCode takes them very seriously. Thank you.

Source: Windows Questions C++