Hello i m reading chapter in K&R about bit operation whish totally confuses me i understand what all operators does but i dont understand the code itself

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

it will right shift p+1-1 then after that i dont understand

~(~0 << n) makes a bitmask of ones equal to the number 'n'. does this by first filling the integer full of ones, then shifting them to the left 'n' bits (filling lsb's with zeros), then inverts all the bits.

for instance, n=5 makes a bitmask = '...00000011111' x >> (p+1-n) shifts the original number to the right 'p+1-n' bits.

the whole point is to extract the bits from the original value, starting with the zero-indexed bit # at 'p' (pointer) and extracting 'n' number of bits moving to the right.

the two are ANDed together to get the new, extracted, value

example value = 0x55 (01010101) with p=6 and n=5 will extract bits 6 through bit 2 (total of 5 bits), to find the new value 0x15, (dec 21, or '10101')

~(~0 << n) makes a bitmask of ones equal to the number 'n'. does this by first filling the integer full of ones, then shifting them to the left 'n' bits (filling lsb's with zeros), then inverts all the bits.

for instance, n=5 makes a bitmask = '...00000011111' x >> (p+1-n) shifts the original number to the right 'p+1-n' bits.

the whole point is to extract the bits from the original value, starting with the zero-indexed bit # at 'p' (pointer) and extracting 'n' number of bits moving to the right.

the two are ANDed together to get the new, extracted, value

example value = 0x55 (01010101) with p=6 and n=5 will extract bits 6 through bit 2 (total of 5 bits), to find the new value 0x15, (dec 21, or '10101')

01010101
p=6 ^
n=5 .....
10101 val = 0x15 (d 21)

but srry 1 more question why would it exract to 2 shoudlnt it exract from 6 till the end ?

to clarify, given my example values (val = 0x55, p=6, n=5), look at the first part of the operation: x >> (p+1-n) ... you notice that it will shift 'x' (the value) a total of two places to the right (6+1-5) = 2.

this drops off the first two bits, becasue the formula is asking for the bits #2 through #6 to be extracted. when that shifted value is ANDed with with the bitmask (...0000011111) it pulls out exactly the five bits remaining at the lowest (rightmost) positions once bits #0 and bits #1 were dropped off.

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