strange behavior of unsigned long long int in loop

Question

While was solving this problem on hackerrank, I noticed a strange thing in the for loop. First, let me show an example code:

#include <bits/stdc++.h>
using namespace std;
#define modVal 1000000007;

int main() {
    for(long long int i=2;i>=0;--i){
        cout<<"here: "<<i<<endl;
    }
}

input: 123

output: here: 2 here: 1 here: 0 164

Now, when I change long long int to unsigned long long int in for loop for the initialization of variable i. The variable i gets initialized with 18446744073709551615. Why is this happening?

it doesn't get initialised with 18446744073709551615. it gets initialised to 2. And because it is unsigned, after two goes round the loop it gets set to a very large positive number. Being unsigned, it can't ever be negative. — Tom Tanner, Jun 25 '19 at 10:29
**Recommended reading:** [Why should I not #include ?](https://stackoverflow.com/q/31816095/560648) — Lightness Races in Orbit, Jun 25 '19 at 10:36

Michael Chourdakis · Accepted Answer · 2019-06-25T11:19:58.387

6

When the variable is unsigned, i >= 0 is always true. So your loop never ends. When i gets to 0, the next -- makes i 0xFFFFFFFFFFFFFFFF (decimal 18446744073709551615).

edited Jun 25 '19 at 11:19

answered Jun 25 '19 at 10:23

Michael Chourdakis

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To be fair, this is a common bug and, although once experienced you look out for it, in a sense it's quite subtle. After all, you told your loop not to go past 0! Sort of. – Lightness Races in Orbit Jun 25 '19 at 11:17
A good compiler however will warn on i >=0 (Condition is always true) – Michael Chourdakis Jun 25 '19 at 12:42
@hs use an IDE. – Michael Chourdakis Jun 25 '19 at 18:50
2

@hs if this answer solved your problem you might want to mark it as the correct answer. Then in a search people can see that the problem has been solved. – findusl Jun 27 '19 at 08:27

Hatted Rooster · Answer 2 · 2019-06-25T10:29:33.323

Because unsigned types can't be negative, attempting to set them to a negative value will make them wrap around and instead hold std::numeric_limits<T>::max() - abs(value) + 1 where T is the type and value the value below 0.

In your loop once i reaches 0 the condition i >= 0 is still met and thus it would get decremented to -1 but that is impossible for unsigned types as explained above and thus the loop will never exit.

Christina Jacob · Answer 3 · 2019-06-25T10:38:19.410

The unsigned numbers as the name suggests don't take signed values. So when i = -1 it is actually 0xFFFFFFFFFFFFFFFF(18446744073709551615 in decimal).

You can see that yourself with the modified program.

 #include <bits/stdc++.h>
 using namespace std;
 #define modVal 1000000007;

 int main() {
    for(unsigned long long int i=2;i>=0;--i){
         cout<<"here: "<<i<<endl;
       if(i > 3)
          return 0;
   }
 }

strange behavior of unsigned long long int in loop

3 Answers3