c# Retrieving INDEX of a bit rather then its VALUE

[SoftwareTester]

I would like to retrieve the INDEX of a bit rather then its VALUE while scanning for MSB or LSB

Example from http://aggregate.org/MAGIC/#Most%20Significant%201%20Bit:
unsigned int msb32(register unsigned int x)
{
x |= (x >> 1);
x |= (x >> 2);
x |= (x >> 4);
x |= (x >> 8);
x |= (x >> 16);
return(x & ~(x >> 1));
}

will not provide the index of a bit (range 0-32) but values from 0 - Maxvalue .

How can i achieve that?

[Plater]

if i read that correctly that function will find the most significant bit that is a "1" and then set all less-significant bits to "1", regardless of what they were prior.
Which if you process that, could tell you which bit was the highest bit with a "1".

I don't see what that has to do with bitwise endian though. As if an 8bit value for the decimal number 10 was:
stored msb->lsb:
0000 1010
or lsb->msb
0101 0000

Regardless of which way it is stored, using the << and >> will have the same result because they too will be setup for that bit order.


Byte-wise endian on the other hand can be easy to check for. Tips from Fortran:
If you have an Int16 and the value is set to decimal 10, then casting it to an 8bit byte, will either give you a 0 or a 10, depending on the order.
If you can cast it to a byte[], you can do it for int32, int64 values as well.

If you really just want the index of the most significant 1s bit:

public static int msb(UInt16 val)
        {
            //16bit val
            //Range:  0 to ((2^16)-1)
            if ((val & 0x8000) > 0)
                return 15;
            if ((val & 0x4000)>0)
                return 14;
            if ((val & 0x2000)>0)
                return 13;
            if ((val & 0x1000)>0)
                return 12;
 
            if ((val & 0x0800)>0)
                return 11;
            if ((val & 0x0400)>0)
                return 10;
            if ((val & 0x0200)>0)
                return 9;
            if ((val & 0x0100)>0)
                return 8;
 
            if ((val & 0x0080)>0)
                return 7;
            if ((val & 0x0040)>0)
                return 6;
            if ((val & 0x0020)>0)
                return 5;
            if ((val & 0x0010)>0)
                return 4;
            if ((val & 0x0008) > 0)
                return 3;
            if ((val & 0x0004) > 0)
                return 2;
            if ((val & 0x0002) > 0)
                return 1;
            if ((val & 0x0001) > 0)
                return 0;
            return -1;
        }
 

I have folded this into a simple for loop before, but could not find the code offhand and didn't want to think about it counting down (i think the magic number is like 17 or something?)