Previous power of 2

Question

There is a lot of information on how to find the next power of 2 of a given value (see refs) but I cannot find any to get the previous power of two.

The only way I f

Answer 1

Here is a one liner for posterity (ruby):

2**Math.log(input, 2).floor(0)

Answer 2

When you work in base 2, you can jump from a power of two to the next one by just adding or removing a digit from the right.

For instance, the previous power of two of the number 8 is the number 4. In binary:

01000 -> 0100 (we remove the trailing zero to get number 4)

So the algorithm to solve the calculus of the previous power of two is:

previousPower := number shr 1

previousPower = number >> 1

(or any other syntax)

Answer 3

This can be done in one line.

int nextLowerPowerOf2 = i <= 0
                        ? 0
                        : ((i & (~i + 1)) == i)
                            ? i >> 1
                            : (1 << (int)Math.Log(i, 2));

result

i    power_of_2
-2    0
-1    0
0    0
1    0
2    1
3    2
4    2
5    4
6    4
7    4
8    4
9    8

Here's a more readable version in c#, with the <=0 guard clause distributed to the utility methods.

int nextLowerPowerOf2 = IsPowerOfTwo(i) 
    ? i >> 1 // shift it right
    : GetPowerOfTwoLessThanOrEqualTo(i);

public static int GetPowerOfTwoLessThanOrEqualTo(int x)
{
    return (x <= 0 ? 0 : (1 << (int)Math.Log(x, 2)));
}

public static bool IsPowerOfTwo(int x)
{
    return (((x & (~x + 1)) == x) && (x > 0));
}

Answer 4

Solution with bit manipulation only:

long FindLargestPowerOf2LowerThanN(long n)
{
    Assert.IsTrue(n > 0);

    byte digits = 0;
    while (n > 0)
    {
        n >>= 1;
        digits++;
    }                            

    return 1 << (digits - 1);
}

Example:

FindLargestPowerOf2LowerThanN(6):
Our Goal is to get 4 or 100
1) 6 is 110
2) 110 has 3 digits
3) Since we need to find the largest power of 2 lower than n we subtract 1 from digits
4) 1 << 2 is equal to 100 

FindLargestPowerOf2LowerThanN(132):
Our Goal is to get 128 or 10000000
1) 6 is 10000100
2) 10000100 has 8 digits
3) Since we need to find the largest power of 2 lower than n we subtract 1 from digits
4) 1 << 7 is equal to 10000000 

Answer 5

From Hacker's Delight, a nice branchless solution:

uint32_t flp2 (uint32_t x)
{
    x = x | (x >> 1);
    x = x | (x >> 2);
    x = x | (x >> 4);
    x = x | (x >> 8);
    x = x | (x >> 16);
    return x - (x >> 1);
}

This typically takes 12 instructions. You can do it in fewer if your CPU has a "count leading zeroes" instruction.