How to efficiently find Bitwise OR of a range of numbers

Given a range of number [a,b], how to efficiently find Bitwise OR of all numbers in this range. Running a loop for range [a,b] and computing Bitwise OR of all the numbers individually is too much time consuming for a range which is very large, so this is not the option.

Instead of doing it for all numbers, you can do it for all positions. That would require you only log(n) steps.

So lets try to imagine - when will units place be 1? If either upper or lower is odd or if there is one number between them. So either lower % 2 == 1 or lower != upper.

Great we got units place. Now if remove the lower one bit from both upper and lower bits and repeat we get the other positions.

Only a special case if lower == upper. In that case we return the lower itself.

Following is the code -

unsigned int bitwiseor(unsigned int a, unsigned int b){
    if (a==b)
        return a;
    unsigned final = 0;
    unsigned rev = 0;
    while(b){
        final*=2;
        if (a%2==1 || a != b)
            final++;
        a/=2;
        b/=2;
    }
    while(final){
        rev *= 2;
        rev += final % 2;
        final/=2;
    }
    return rev;
}

The second loop is to just reserve the bit sequence.

Demo here - https://ideone.com/MCIugW

Thank you @Meixner for the driver code.