Unsigned vs Signed Integer [duplicate] - c

Question

This question already has answers here:
Why does counting down an unsigned int loop forever in C?
(5 answers)
Closed 4 years ago.
What's wrong with this code?
#include<stdio.h>
int main()
{
unsigned int i;
for(i=100;i>=0;i--)
{
printf("%u ",i);
}
return 0;
}
This code doesn't work but if i use for(i=100;i>0;i--) then it works!
or another way is to use integer instead of using unsigned integer.

Answer 1

An unsigned int can never be negative, so i >= 0 holds true all the time. So, it is effectively an infinite loop.

Answer 2

Since variable i is declared as unsigned int, this condition i>=0 never fails, hence results in infinite loop rotation.
unsigned int i;
for(i=100;i>=0;i--) { /* i will never become negative */
printf("%u \n",i);
}
Note that UINT_MAX is 4294967295 see this i.e it ranges from 0 to 4294967295 so when i=0 it prints & gets decremented, next it won't be -1, it will be 4294967295, hence above for loop results in infinite loop.
another way is to use integer instead of using unsigned integer ? you can do the same using unsigned integer also by replacing the condition part as i>0 so that when i=0 it fails & comes out of loop.

Answer 3

What's wrong with this code?
i>=0 is always true as i is unsigned. Then for(i=100;i>=0;i--) loops forever.
To continue using unsigned i:
Since the coding goal always wants to enter the loop at least once, instead of testing the loop condition in the begining, test at the end.
//for(i=100;i>=0;i--) {
// printf("%u ",i);
//}
i = 100; // or any unsigned constant.
do {
printf("%u ",i);
} while (i-- > 0);
another way is to use integer (int) instead of using unsigned integer (unsigned).(?)
Use int. Of course this will not work if i needs to start at values greater than INT_MAX. If such large values are needed, consider a wider signed type like long long.
int i;
for(i=100;i>=0;i--) {
printf("%u ",i);
}