Swapping bits at a given point between two bytes - bit-manipulation

Question

Let's say I have these two numbers:
x = 0xB7
y = 0xD9
Their binary representations are:
x = 1011 0111
y = 1101 1001
Now I want to crossover (GA) at a given point, say from position 4 onwards.
The expected result should be:
x = 1011 1001
y = 1101 0111
Bitwise, how can I achieve this?

Answer 1

I would just use bitwise operators:
t = (x & 0x0f)
x = (x & 0xf0) | (y & 0x0f)
y = (y & 0xf0) | t
That would work for that specific case. In order to make it more adaptable, I'd put it in a function, something like (pseudo-code, with &, | and ! representing bitwise "and", "or", and "not" respectively):
def swapBits (x, y, s, e):
lookup = [255,127,63,31,15,7,3,1]
mask = lookup[s] & !lookup[e]
t = x & mask
x = (x & !mask) | (y & mask)
y = (y & !mask) | t
return (x,y)
The lookup values allow you to specify which bits to swap. Let's take the values xxxxxxxx for x and yyyyyyyy for y along with start bit s of 2 and end bit e of 6 (bit numbers start at zero on the left in this scenario):
x y s e t mask !mask execute
-------- -------- - - -------- -------- -------- -------
xxxxxxxx yyyyyyyy 2 6 starting point
00111111 mask = lookup[2](00111111)
00111100 & !lookup[6](11111100)
00xxxx00 t = x & mask
xx0000xx x = x & !mask(11000011)
xxyyyyxx | y & mask(00111100)
yy0000yy y = y & !mask(11000011)
yyxxxxyy | t(00xxxx00)

Answer 2

If a bit position is the same in both values, no change is needed in either. If it's opposite, they both need to invert.
XOR with 1 flips a bit; XOR with 0 is a no-op.
So what we want is a value that has a 1 everywhere there's a bit-difference between the inputs, and a 0 everywhere else. That's exactly what a XOR b does.
Simply mask this bit-difference to only keep the differences in the bits we want to swap, and we have a bit-swap in 3 XORs + 1 AND.
Your mask is (1UL << position) -1. One less than a power of 2 has all the bits below that set. Or more generally with a high and low position for your bit-range: (1UL << highpos) - (1UL << lowpos). Whether a lookup-table is faster than bit-set / sub depends on the compiler and hardware. (See #PaxDiablo's answer for the LUT suggestion).
// Portable C:
//static inline
void swapBits_char(unsigned char *A, unsigned char *B)
{
const unsigned highpos = 4, lowpos=0; // function args if you like
const unsigned char mask = (1UL << highpos) - (1UL << lowpos);
unsigned char tmpA = *A, tmpB = *B; // read into locals in case A==B
unsigned char bitdiff = tmpA ^ tmpB;
bitdiff &= mask; // clear all but the selected bits
*A = tmpA ^ bitdiff; // flip bits that differed
*B = tmpB ^ bitdiff;
}
//static inline
void swapBit_uint(unsigned *A, unsigned *B, unsigned mask)
{
unsigned tmpA = *A, tmpB = *B;
unsigned bitdiff = tmpA ^ tmpB;
bitdiff &= mask; // clear all but the selected bits
*A = tmpA ^ bitdiff;
*B = tmpB ^ bitdiff;
}
(Godbolt compiler explorer with gcc for x86-64 and ARM)
This is not an xor-swap. It does use temporary storage. As #chux's answer on a near-duplicate question demonstrates, a masked xor-swap requires 3 AND operations as well as 3 XOR. (And defeats the only benefit of XOR-swap by requiring a temporary register or other storage for the & results.) This answer is a modified copy of my answer on that other question.
This version only requires 1 AND. Also, the last two XORs are independent of each other, so total latency from inputs to both outputs is only 3 operations. (Typically 3 cycles).
For an x86 asm example of this, see this code-golf Exchange capitalization of two strings in 14 bytes of x86-64 machine code (with commented asm source)

Answer 3

Swapping individual bits with XOR
unsigned int i, j; // positions of bit sequences to swap
unsigned int n; // number of consecutive bits in each sequence
unsigned int b; // bits to swap reside in b
unsigned int r; // bit-swapped result goes here
unsigned int x = ((b >> i) ^ (b >> j)) & ((1U << n) - 1); // XOR temporary
r = b ^ ((x << i) | (x << j));