Dennis Ritchie doubt pg 49

Question

As an illustration of some of the bit operators, consider the function getbits(x,p,n) that returns the (right adjusted) n-bit field of x that begins at position p. We assume that bit position 0 is at the right end and that n and p are sensible positive values. For example, getbits(x,4,3) returns the three bits in positions 4, 3 and 2, right-adjusted.

Code:

/* getbits: get n bits from position p */
    unsigned getbits(unsigned x, int p, int n)
    {
         return (x >> (p+1-n)) & ~(~0 << n);
    }

The expression x >> (p+1-n) moves the desired field to the right end of the word. ~0 is all 1-bits; shifting it left n positions with ~0<<n places zeros in the rightmost n bits; complementing that with ~ makes a mask with ones in the rightmost n bits.

Can anyone please explain this to me??

What's happening here???

Answer 1

getbits(x,p,n) works like get n bit field of x from position p 

Consider getbits(4,3,2) 

Here, x = 4, p = 3 and n = 2

It is also mentioned that bit position 0 is at the right end.

x = 0000 0100 

So, by definition it should return 2 bits value from position 3 i.e. 01 whose value is 1.

Now, let's perform 

return (x >> (p+1-n)) & ~(~0 << n);

p + 1 - n = 3 + 1 - 2 = 2

So, x is right shifted by 2. Hence x becomes 0000 0001     ....(1)

~0 = 1111 1111 is left shifted by n i.e. 2

(~0 << n) = 1111 1100

and finally ~(~0 << n) = 0000 0011   ....(2)

Performing AND operation of (1) and (2) , we get 0000 0001 whose value is 1.

Basically ~(~0 << n) is acting like a mask of all 1's to get the right adjusted n bit field from position p of x. 

Comments

Thanks....:)

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Can u Plzz explain masking...plzz

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Refer this https://stackoverflow.com/questions/10493411/what-is-bit-masking

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Thanks again....