for (unsigned char i = 0; i<=0xff; i++) produces infinite loop

Question

Why does the following c code end up in an infinite loop?

for(unsigned char i = 0; i <= 0xff; i++){}

It is the same result with:

Answer 1

Because you wraparound. An unsigned char values range between 0-255, and because unsigned arithmetic are well defind, you actually wrap the value of i to 0 and the condition is still met and the iteration continues ad infinitum.

In the case of signed values, it is an undefined behaviour and value stored in i might not be 0. But still it will be smaller than the maximum value you could store in a char and the condition is still met.