Finding Bit Positions in an unsigned 32-bit integer

Question

I think I might have been asleep in my CS class when they talked about Bit Positions, so I am hoping someone can lend a hand.

I have a unsigned 32-bit integer (Lets use the value: 28)

According to some documentation I am going over, the value of the integer contains flags specifying various things.

Bit positions within the flag are numbered from 1 (low-order) to 32 (high-order). All undefined flag bits are reserved and must be set to 0.

I have a Table that shows the meanings of the flags, with meaning for the numbers 1-10.

So my question is... WTF? :-)

I am hoping that someone can try and explain to me what this all means and how to find the "flag" value(s) from a number like, 28, based off of bit position.

Thanks

Answer 1

Assuming flags is unsigned...

int flag_num = 1;
while (flags != 0)
{
    if ((flags&1) != 0)
    {
        printf("Flag %d set\n", flags);
    }
    flags >>= 1;
    flag_num += 1;
}

If flags is signed you should replace

flags >>= 1;

with

flags = (flags >> 1) & 0x7fffffff;

Answer 2

28 converts to 11100 in binary. That means bits 1 and 2 are not set and bits 3, 4 and 5 are set.

A few points: first, anybody who's really accustomed to C will usually start the numbering at 0, not 1. Second, you can test of individual flags with the bitwise and operator (&), as in:

#define flag1 1    //  1 = 00 0001
#define flag2 2    //  2 = 00 0010
#define flag3 4    //  4 = 00 0100
#define flag4 8    //  8 = 00 1000
#define flag5 16   // 16 = 01 0000
#define flag6 32   // 32 = 10 0000

if (myvalue & flag1)
    // flag1 was set

if (myvalue & flag4)
    // flag4 was set

and so on. You can also check which bits are set in a loop:

#include <stdio.h>

int main() { 
    int myvalue = 28;
    int i, iter;

    for (i=1, iter=1; i<256; i<<=1, iter++)
        if (myvalue & i)
            printf("Flag: %d set\n", iter);
    return 0;
}

should print:

Flag: 3 set
Flag: 4 set
Flag: 5 set

Answer 3

To get an int with the value 0 or 1 representing just the nth bit from that integer, use:

int bitN = (value >> n) & 1;

But that's not usually what you want to do. A more common idiom is this:

int bitN = value & (1 << n);

In this case bitN will be 0 if the nth bit is not set, and non-zero in the case that the nth bit is set. (Specifically, it'll be whatever value comes out with just the nth bit set.)