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 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 ", 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)…

Accepted Answer

`((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
```

Bit invert function in K&R exercise 2-7

Solution

Problem