Mask and extract bits in C

Question

I've been looking at posts about mask but still can't get my head around how to extract certain bits from a number in C.

Say if we have a int number 0001 1010 0100 1011 , so it's hex representation is x1a4b right? If I want to know the 5th to 7th number, which is 101 in this case, shall I use int mask= 0x0000 1110 0000 0000, int extract = mask&number?

Also how can I check if it is 101? I guess == won't work here... Many thanks!

Answer 1

Masking is done by setting all the bits except the one(s) you want to 0. So let's say you have a 8 bit variable and you want to check if the 5th bit from the is a 1. Let's say your variable is 00101100. To mask all the other bits we set all the bits except the 5th one to 0 using the & operator:

00101100 & 00010000

Now what this does is for every bit except the 5th one, the bit from the byte on the right will be 0, so the result of the & operation will be 0. For the 5th bit, however, the value from the right bit is a 1, so the result will be whatever the value of hte 5th bit from the left byte is - in this case 0:

Now to check this value you have to compare it with something. To do this, simply compare the result with the byte on the right:

result = (00101100 & 00010000) == 00000000

To generalize this, you can retrieve any bit from the lefthand byte simply by left-shifting 00000001 until you get the bit you want. The following function achieves this:

int getBit(char byte, int bitNum)
{
    return (byte & (0x1 << (bitNum - 1)))
}

This works on vars of any size, whether it's 8, 16, 32 or 64 (or anything else for that matter).

Answer 2

Assuming the gcc extension 0b to define binary literals:

int number = 0b0001101001001011; /* 0x1a4b */
int mask =   0b0000111000000000; /* 0x0e00 */
/* &'ed:     0b0000101000000000;    0x0a00 */
int extract = mask & number;     /* 0x0a00 */

if (extract == 0b0000101000000000)
/* or if 0b is not available:
if (extract == 0x0a00 ) */
{
  /* success */
}
else
{
  /* failure */
}