Bit Reversal using bitwise

Question

Hi all

I am trying to do bit reversal in a byte. I use the code below

static int BitReversal(int n)
{
    int u0 = 0x55555555; // 01010101010101010101010101010101
    int u1 = 0x33333333; // 00110011001100110011001100110011
    int u2 = 0x0F0F0F0F; // 00001111000011110000111100001111
    int u3 = 0x00FF00FF; // 00000000111111110000000011111111
    int u4 = 0x0000FFFF;
    int x, y, z;
    x = n;
    y = (x >> 1) & u0;
    z = (x & u0) << 1;
    x = y | z;

    y = (x >> 2) & u1;
    z = (x & u1) << 2;
    x = y | z;

    y = (x >> 4) & u2;
    z = (x & u2) << 4;
    x = y | z;

    y = (x >> 8) & u3;
    z = (x & u3) << 8;
    x = y | z;

    y = (x >> 16) & u4;
    z = (x & u4) << 16;
    x = y | z;

    return x;
}

It can reverser the bit (on a 32-bit machine), but there is a problem, For example, the input is 10001111101, I want to get 10111110001, but this method would reverse the whole byte including the heading 0s. The output is 10111110001000000000000000000000. Is there any method to only reverse the actual number? I do not want to convert it to string and reverser, then convert again. Is there any pure math method or bit operation method?

Best Regards,

Answer 1

Get the highest bit number using a similar approach and shift the resulting bits to the right 33 - #bits and voila!

Answer 2

One method could be to find the leading number of sign bits in the number n, left shift n by that number and then run it through your above algorithm.

Answer 3

Cheesy way is to shift until you get a 1 on the right:

if (x != 0) {
    while ((x & 1) == 0) {
        x >>= 1;
    }
}

Note: You should switch all the variables to unsigned int. As written you can have unwanted sign-extension any time you right shift.

Answer 4

It's assuming all 32 bits are significant and reversing the whole thing. You COULD try to make it guess the number of significant bits by finding the highest 1, but that isn't necessarily accurate so I'd suggest you modify the function so it takes a second parameter indicating the number of significant bits. Then after reversing the bits just shift them to the right.

Answer 5

Try using Integer.reverse(int x);