Need help understanding "getbits()" method in Chapter 2 of K&R C

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.

I would say the best thing to do is to do a problem out by hand, that way you'll understand how it works.

Here is what I did using an 8-bit unsigned int.

1. Our number is 75 we want the 4 bits starting from position 6. the call for the function would be getbits(75,6,4);

2. 75 in binary is 0100 1011

3. So we create a mask that is 4 bits long starting with the lowest order bit this is done as such.

~0 = 1111 1111
<<4 = 1111 0000
~ = 0000 1111

Okay we got our mask.

4. Now, we push the bits we want out of the number into the lowest order bits so we shift binary 75 by 6+1-4=3.

0100 1011 >>3 0000 1001

Now we have a mask of the correct number of bits in the low order and the bits we want out of the original number in the low order.

4. so we & them
   

  0000 1001 

& 0000 1111
============

  0000 1001

so the answer is decimal 9.

Note: the higher order nibble just happens to be all zeros, making the masking redundant in this case but it could have been anything depending on the value of the number we started with.

~(~0 << n) creates a mask that will have the n right-most bits turned on.

0
   0000000000000000
~0
   1111111111111111
~0 << 4
   1111111111110000
~(~0 << 4)
   0000000000001111

ANDing the result with something else will return what's in those n bits.

Edit: I wanted to point out this programmer's calculator I've been using forever: AnalogX PCalc.