addition at the left of number


How we can we do addition at the left FF instead 00? For example we have a = E8, we need a = 0xFFFFFFe8

 0xFFE26C02 -> 0xFFE26C02
 0x000000e8 -> 0xFFFFFFe8
 0x100000e8 -> 0x100000e8
 0x001000e8 -> 0xFF1000e8

P.S. data type int32 or int64

asked on Stack Overflow Dec 12, 2013 by Qwerty • edited Dec 12, 2013 by Qwerty

3 Answers


I guess your problem is how put the bits left to the leftmost '1' in the number?

Here is my solution:

Let's take the number 01001000( as num ) for example:

  1. First, do ~num and you get 10110111.
  2. Then plus 1, and the answer of (~num+1) is 10111000.
  3. Do (~num+1) & (num), and you get 00001000.

So in the end, the result of the (~num+1) & num is to get the rightmost '1' in the num.

To get the left most one, do a loop, and the condition is:


In the end, you will get the number 01000000.

So it'll be easy to locate the '0' right to the leftmost '1'.

Hope it will help.

answered on Stack Overflow Dec 12, 2013 by VELVETDETH

You can use the following to convert all leading bytes of 0 to FF

int RevLeadingZeros(int number)
    if((number & 0xFF000000)==0)
        number |= 0xFF000000
        return number;
    if((number & 0x00FF0000)==0)
        number |= 0x00FF0000
        return number;
    if((number & 0x0000FF00)==0)
        number |= 0x0000FF00
        return number;
    if((number & 0x000000FF)==0)
        number |= 0x000000FF
        return number;
answered on Stack Overflow Dec 12, 2013 by Dipto • edited Dec 12, 2013 by Raxvan

Without direct calculation, you can just iterate each byte of your integer and replace the last consecutive 00 with FF.

You can also OR your integer with a mask which can easily be built with a first traversing of your integer memory.

answered on Stack Overflow Dec 12, 2013 by Oragon Efreet

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