Question

In chapter 2, the section on bitwise operators (section 2.9), I’m having trouble understanding how one of the sample methods works.

Here’s the method provided:

unsigned int getbits(unsigned int x, int p, int n) {     return (x >> (p + 1 - n)) & ~(~0 << n); } 

The idea is that, for the given number x, it will return the n bits starting at position p, counting from the right (with the farthest right bit being position 0). Given the following main() method:

int main(void) {     int x = 0xF994, p = 4, n = 3;     int z = getbits(x, p, n);     printf('getbits(%u (%x), %d, %d) = %u (%X)\n', x, x, p, n, z, z);      return 0; } 

The output is:

getbits(63892 (f994), 4, 3) = 5 (5)

I get portions of this, but am having trouble with the ‘big picture,’ mostly because of the bits (no pun intended) that I don’t understand.

The part I’m specifically having issues with is the complements piece: ~(~0 << n). I think I get the first part, dealing with x; it’s this part (and then the mask) that I’m struggling with — and how it all comes together to actually retrieve those bits. (Which I’ve verified it is doing, both with code and checking my results using calc.exe — thank God it has a binary view!)

Any help?

Answer 1

Let’s use 16 bits for our example. In that case, ~0 is equal to

1111111111111111 

When we left-shift this n bits (3 in your case), we get:

1111111111111000 

because the 1s at the left are discarded and 0s are fed in at the right. Then re-complementing it gives:

0000000000000111 

so it’s just a clever way to get n 1-bits in the least significant part of the number.

The "x bit" you describe has shifted the given number (f994 = 1111 1001 1001 0100) right far enough so that the least significant 3 bits are the ones you want. In this example, the input bits you’re requesting are there, all other input bits are marked . since they’re not important to the final result:

ff94             ...........101..  # original number >> p+1-n     [2] .............101  # shift desired bits to right & ~(~0 << n) [7] 0000000000000101  # clear all the other (left) bits 

As you can see, you now have the relevant bits, in the rightmost bit positions.