Bit invert function in K&R exercise 2-7

Question

Exercise 2-7 of The C Programming Language:

Write a function invert(x,p,n) that returns x with the n bits that begin at position p inverted (i.e., 1 changed to 0 and vice versa), leaving the others unchanged.

I understood the question like this: I have 182 which is 101(101)10 in binary, the part in parentheses has to be inverted without changing the rest. The return value should be 10101010 then, which is 170 in decimal.

Here is my attempt:

#include <stdio.h>

unsigned int getbits(unsigned int bitfield, int pos, int num);
unsigned int invert(unsigned int bitfield, int pos, int num);

int main(void)
{
    printf("%d\n", invert(182, 4, 3));
    return 0;
}

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

/* invert: flip pos-num bits in bitfield */
unsigned int invert(unsigned int bitfield, int pos, int num)
{
    unsigned int mask;
    unsigned int bits = getbits(bitfield,pos,num);

    mask = (bits << (num-1)) | ((~bits << (pos+1)) >> num);

    return bitfield ^ mask;
}

It seems correct (to me), but invert(182, 4, 3) outputs 536870730. getbits() works fine (it's straight from the book). I wrote down what happens in the expression I've assigned to y:

(00000101 << 2) | ((~00000101 << 5) >> 3)    -- 000000101 is the part being flipped: 101(101)10
       00010100 | ((11111010 << 5) >> 3)
       00010100 | (01000000 >> 3)
       00010100 | 00001000
= 00011100

  10110110 (182)
^ 00011100
----------
= 10101010 (170)

Should be correct, but it isn't. I found out this is where it goes wrong: ((~xpn << (p+1)) >> n). I don't see how.

Also, I've no idea how general this code is. My first priority is to just get this case working. Help in this issue is welcome too.

Answer 1

((1u<<n)-1) is a bit mask with n '1' bits at the RHS. <<p shifts this block of ones p positions to the left. (you should shift with (p-n) instead of p if you want to count from the left).

return val  ^ (((1u<<n)-1) <<p) ;

There still is a problem when p is larger than the wordsize (shifting by more than the wordsize is undefined), but that should be the responsability of the caller ;-)

For the example 101(101)10 with p=2 and n=3:

1u<<n               := 1000
((1u<<n)-1)         := 0111
(((1u<<n)-1) <<p) := 011100
original val    := 10110110
val ^ mask      := 10101010