[SOLVED] Why is the code part infinite?

Question

This is an infinite loop.
for(unsigned i=0; i >=0; i--) {
// do work ...
}

Answer 1

That’s because an unsigned int can only be a positive number.

i already equals 0, it cannot go into negatives. i will always be greater than or equal to zero, so the check i >= 0 will always return true.

It’s effectively the same as:
for (;;).

Here’s an example that is not infinite, using a signed int which allows negatives:

for (int i = 0; i >= -10; --i)
{
printf("%d", i);

}e

Abhay explains below, that on the decrement iteration where i equals zero, the counter goes back to the max positive value, like an odometer.

Answer 2

It does not only remains zero as i thought . Look at the example ,

for(unsigned int i=0;i>=0;i--){
printf("%lu\n",i);
}

It will start decrementing from the max value of unsigned int after we try to make the number negative .

Answer 3

As i is unsigned data type so its value never become negative so it becomes infinite

Answer 4

Thanks for all explanations! :)
DavX ,
Abhay ,
Mohd Aadil .