Question

We’ve two numbers with same bit patterns in their lower order.
For ex: 01001110110 and 10110 are the two numbers, they match with their lower order.

Is there a simple way to find this out ?
I’ve a solution with shifting the bits and then comparing, Is there a better way ?

Answer 1

You can XOR them together and check if the last N lower order bits are all zero (where N is the number of bits in the smaller of the two numbers).

For eg: using the sample numbers you gave, 01001110110 and 10110:

01001110110 XOR 10110 = 01001100000

Notice that the last 5 bits are all zero in the result.

In C/C++/Java you can use the ^ operator for this purpose and then extract the last N bits with a mask like so:

int a = 0x276; // 01001110110
int b = 0x16;  //       10110

if (((a ^ b) & 0x1F) == 0) { // Mask 0x1F assumes least significant 5 bits for match
    // match!
}

If course, this assumes you know the number of significant bits in each number (5 in this example). If instead the number of matching bits is unspecified, you will need to count the number of consecutive trailing 0s to figure out how many bits match. There may be some other trickery you could perform in this case.