Code for displaying the smallest power of 2 greater than or equal than the integer

Question

I need a code in Java that can find the smallest power of 2 greater than or equal to any non-negative integer entered by the user. Can anyone help?

Eclipse is an IDE what language? Plus this smells of a homework question. What have you tried? — David Basarab, Sep 23 '09 at 00:19
Sorry it asked me to tag it with something so I just chose Eclipse. The language is Java and I'm honestly not sure of where to begin on it aside from prompting for an interger and making sure that it is postive. — , Sep 23 '09 at 00:25

score 7 · Answer 1 · answered Sep 23 '09 at 01:36

7

i>1 ? Integer.highestOneBit(i-1)<<1 : 1

Obviously suffers from integer overflow (there isn't a strictly correct solution in int for around half of positive ints).

Usual disclaimer: Not tested or compiled.

answered Sep 23 '09 at 01:36

Tom Hawtin - tackline

145,806
30
211
305

Why not use Long.highestOneBit? – Yishai Sep 23 '09 at 01:53
Well, you could do, but if you started with an `int`, you probably want an `int` result anyway. Also you can't do that for `long`. Arbitrary precision integers ftw! – Tom Hawtin - tackline Sep 23 '09 at 02:56

score 2 · Answer 2 · edited May 23 '17 at 12:08

2

See this link

Algorithm for finding the smallest power of two that’s greater or equal to a given value

Bye.

edited May 23 '17 at 12:08

Community

1
1

answered Sep 23 '09 at 00:21

RRUZ

134,889
20
356
483

score 0 · Answer 3 · answered Nov 24 '15 at 06:41

0

int nextPow2(int n) {
  if (n <= 1) return n;
  double d = n - 1;
  return 1 << ((Double.doubleToRawLongBits(d) >> 52) - 1022);
}

Fastest solution I have found on desktop.

answered Nov 24 '15 at 06:41

hoford

4,918
2
19
19

score 0 · Answer 4 · answered Nov 25 '18 at 16:14

Smallest power of 2 greater than or equal to a

     // Next higher power of 2 greater than or equal to a
     public static int nextPow2(int a) {
            int r0, r1, r2, r3, r4;
            r0 = 2 * highestOneBit(a - 1);
            r1 = highestOneBit(a - 1) << 1;
            r2 = (int) pow(2, ceil(log(a) / log(2)));
            r3 = (int) pow(2, ceil(log10(a) / log10(2)));
            r4 = (int) pow(2, 32 - numberOfLeadingZeros(a - 1));
            return r0; // or r1 or r2 or r3 or r4
        }

Exponent of the smallest power of 2 greater than or equal to a

    // Exponent of next higher power of 2 greater than or equal to a
    public static int nextpow2(int a) {
        int r0, r1, r2;
        r0 = (int) ceil(log(a) / log(2));
        r1 = (int) ceil(log10(a) / log10(2));
        r2 = (int) 32 - numberOfLeadingZeros(a - 1); // ceil(log2(x)) = 32 - numberOfLeadingZeros(x - 1)
        return r0; // or r1 or r2
    }

score -1 · Answer 5 · answered Sep 23 '09 at 01:26

-1

if (i==0) return 1;
else if (i==1) return 2;
else if (i==2 || i==3) return 4;
else if (i==4 || i==5 || i==6 || i==7) return 8;
else if (...)

answered Sep 23 '09 at 01:26

mob

117,087
18
149
283

score -2 · Answer 6 · answered Sep 23 '09 at 00:29

-2

you can do this by counting the number of unsigned right shifts until the result is zero

answered Sep 23 '09 at 00:29

Scott Evernden

39,136
15
78
84

Code for displaying the smallest power of 2 greater than or equal than the integer

6 Answers6