Power of two failed to pass the test

Question

Isn't this one way to test whether a number is a power of 2?

boolean isPowerOfTwo(int n) {
        return (n%2==0);
    }

This tells you if a number is divisible by 2, not a power of 2 — Carcigenicate, Nov 13 '16 at 16:17
You're checking if the number is even. This will say 6,12, 18,.... are power of 2. — Bijay Gurung, Nov 13 '16 at 16:17
do you think 40 is power of 2? when 40 can be divided by 2?? — ΦXocę 웃 Пepeúpa ツ, Nov 13 '16 at 16:17
this is test if number is `even`. To test for power of 2, read this (or use google) : http://stackoverflow.com/questions/19383248/find-if-a-number-is-a-power-of-two-without-math-function-or-log-function — rkosegi, Nov 13 '16 at 16:17

Soner from The Ottoman Empire · Accepted Answer · 2016-11-13T18:09:02.503

1

Are you sure that you are checking power of 2? Rather, being checked divisible with 2. If you are serious, use the piece of cake snippet

return n > 0 && ((n & -n) == n);

The second way,

private static boolean powerOf2(int num)
{
  if(num <= 0){
   return false;
  }

  while(num > 1)
  {
   if(num % 2 != 0)
   {
    return false;
   }
   num = num / 2;
  }
  return true;
}

edited Nov 13 '16 at 18:09

answered Nov 13 '16 at 16:17

Soner from The Ottoman Empire

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let me come up with some other way, i hate bit operation. – ling Nov 13 '16 at 16:22
Note: this returns true for 0. – Peter Lawrey Nov 13 '16 at 17:21

score 1 · Answer 2 · answered Nov 13 '16 at 16:22

1

That function checks whether or not an integer is even (i.e divisible by 2), not whether or not the integer is a power of 2.

To check if a number is a power of 2, you should look at its bits. ints that are a power of 2 will have one bit set to 1 and all others set to 0.

boolean isPowerOf2(int i) {
    return Integer.bitCount(i) == 1;
}

answered Nov 13 '16 at 16:22

The SE I loved is dead

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Tested, simple, yet inefficient. even worse than the recursive version. – ling Nov 13 '16 at 16:41

score 0 · Answer 3 · answered Nov 13 '16 at 16:17

0

The % is the modulo operator, meaning the remainder after division. your method checks if the int n is divisible by 2

answered Nov 13 '16 at 16:17

GerritCap

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So my code is a necessary, but insufficient condition of being as a power of 2. right? – ling Nov 13 '16 at 16:22
I would not put it in the "necessary" category, you are just checking if the number is dividable by 2 not if it is a power of 2. It is just doing something different. There is however a binary rule: if the binary version of your int only has one 1 digit, then it is a power of 2. This can be checked with a boolean operator n & (n - 1) == 0 – GerritCap Nov 13 '16 at 16:33
i mean it must be divisible by 2, which is necessary. – ling Nov 13 '16 at 16:40
@ling Powers of 2 are actually a subset of even numbers - if you figure out that an integer is a power of 2, you don't need to check if the same integer is even. – The SE I loved is dead Nov 13 '16 at 16:57

Power of two failed to pass the test

3 Answers3