问题
I'm trying to duplicate an 8-bit value to 32-bit and wanted to ask if it's possible to write a single line algorithm to duplicate the bit values.
For example:
1100 1011 -> 1111 1111 0000 0000 1111 0000 1111 1111
If it's possible, I would like to understand what's the logic behind it.
回答1:
It's simple - solve the simplest case, then do more complex ones.
You just need to spread bits by inserting 3 zero bits between then. Once this is done, the final step is:
x = (x << 0) | (x << 1) | (x << 2) | (x << 3);
Or in a less obvious but faster way:
x = (x << 4) - x;
Case 1: Duplicate 2 bits into an 8 bit value (the simplest).
+---+---------+---------+
| 0 | _ _ _ _ | _ _ A B |
+---+---------+---------+
| 1 | _ _ _ A | _ _ _ B |
+---+---------+---------+
| 2 | A A A A | B B B B |
+---+---------+---------+
Case 2: Duplicate 4 bits into a 16 bit value. How? Just move 2 bits to the upper part to turn it into the case 1! Divide and conquer!
+---+---------+---------+---------+---------+
| 0 | _ _ _ _ | _ _ _ _ | _ _ _ _ | A B C D |
+---+---------+---------+---------+---------+
| 1 | _ _ _ _ | _ _ A B | _ _ _ _ | _ _ C D |
+---+---------+---------+---------+---------+
| 2 | _ _ _ A | _ _ _ B | _ _ _ C | _ _ _ D |
+---+---------+---------+---------+---------+
| 3 | A A A A | B B B B | C C C C | D D D D |
+---+---------+---------+---------+---------+
Case 3: Duplicate 8 bits into a 32 bit value (the original).
+---+---------+---------+---------+---------+---------+---------+---------+---------+
| 0 | _ _ _ _ | _ _ _ _ | _ _ _ _ | _ _ _ _ | _ _ _ _ | _ _ _ _ | A B C D | E F G H |
+---+---------+---------+---------+---------+---------+---------+---------+---------+
| 1 | _ _ _ _ | _ _ _ _ | _ _ _ _ | A B C D | _ _ _ _ | _ _ _ _ | _ _ _ _ | E F G H |
+---+---------+---------+---------+---------+---------+---------+---------+---------+
| 2 | _ _ _ _ | _ _ A B | _ _ _ _ | _ _ C D | _ _ _ _ | _ _ E F | _ _ _ _ | _ _ G H |
+---+---------+---------+---------+---------+---------+---------+---------+---------+
| 3 | _ _ _ A | _ _ _ B | _ _ _ C | _ _ _ D | _ _ _ E | _ _ _ F | _ _ _ G | _ _ _ H |
+---+---------+---------+---------+---------+---------+---------+---------+---------+
| 4 | A A A A | B B B B | C C C C | D D D D | E E E E | F F F F | G G G G | H H H H |
+---+---------+---------+---------+---------+---------+---------+---------+---------+
Can be achieved by the code below:
uint32_t interleave(uint8_t value)
{
uint32_t x = value;
x = (x | (x << 12)) /* & 0x000F000F */; // GCC is not able to remove redundant & here
x = (x | (x << 6)) & 0x03030303;
x = (x | (x << 3)) & 0x11111111;
x = (x << 4) - x;
return x;
}
Some test cases to check that it works:
TEST_F(test, interleave)
{
EXPECT_EQ(interleave(0x00), 0x00000000);
EXPECT_EQ(interleave(0x11), 0x000F000F);
EXPECT_EQ(interleave(0x22), 0x00F000F0);
EXPECT_EQ(interleave(0x33), 0x00FF00FF);
EXPECT_EQ(interleave(0x44), 0x0F000F00);
EXPECT_EQ(interleave(0x55), 0x0F0F0F0F);
EXPECT_EQ(interleave(0x66), 0x0FF00FF0);
EXPECT_EQ(interleave(0x77), 0x0FFF0FFF);
EXPECT_EQ(interleave(0x88), 0xF000F000);
EXPECT_EQ(interleave(0x99), 0xF00FF00F);
EXPECT_EQ(interleave(0xAA), 0xF0F0F0F0);
EXPECT_EQ(interleave(0xBB), 0xF0FFF0FF);
EXPECT_EQ(interleave(0xCC), 0xFF00FF00);
EXPECT_EQ(interleave(0xDD), 0xFF0FFF0F);
EXPECT_EQ(interleave(0xEE), 0xFFF0FFF0);
EXPECT_EQ(interleave(0xFF), 0xFFFFFFFF);
EXPECT_EQ(interleave(0x01), 0x0000000F);
EXPECT_EQ(interleave(0x23), 0x00F000FF);
EXPECT_EQ(interleave(0x45), 0x0F000F0F);
EXPECT_EQ(interleave(0x67), 0x0FF00FFF);
EXPECT_EQ(interleave(0x89), 0xF000F00F);
EXPECT_EQ(interleave(0xAB), 0xF0F0F0FF);
EXPECT_EQ(interleave(0xCD), 0xFF00FF0F);
EXPECT_EQ(interleave(0xEF), 0xFFF0FFFF);
}
回答2:
There are only 256 8-bit values, so a simple lookup table would occupy 1kb, and the lookup is trivial. It's hard to believe that any bithack would have superior performance.
回答3:
This would work:
unsigned int eToTW (unsigned char a) {
unsigned int output = 0;
output |= a & 0x80 ? ((unsigned) 0xf) << 28 : 0x0;
output |= a & 0x40 ? 0xf << 24 : 0x0;
output |= a & 0x20 ? 0xf << 20 : 0x0;
output |= a & 0x10 ? 0xf << 16 : 0x0;
output |= a & 0x8 ? 0xf << 12 : 0x0;
output |= a & 0x4 ? 0xf << 8 : 0x0;
output |= a & 0x2 ? 0xf << 4 : 0x0;
output |= a & 0x1 ? 0xf : 0x0;
return output;
}
or this:
unsigned int eToTW (unsigned char a) {
unsigned int output = 0;
output |= a & (1 << 7) ? ((unsigned) 0xf) << 28 : 0x0;
output |= a & (1 << 6) ? 0xf << 24 : 0x0;
output |= a & (1 << 5) ? 0xf << 20 : 0x0;
output |= a & (1 << 4) ? 0xf << 16 : 0x0;
output |= a & (1 << 3) ? 0xf << 12 : 0x0;
output |= a & (1 << 2) ? 0xf << 8 : 0x0;
output |= a & (1 << 1) ? 0xf << 4 : 0x0;
output |= a & 1 ? 0xf : 0x0;
return output;
}
yet another solution:
unsigned int eToTW (unsigned char a) {
return (a & 1 << 7 ? ((unsigned) 0xf) << 28 : 0x0) |
(a & 1 << 6 ? 0xf << 24 : 0x0) |
(a & 1 << 5 ? 0xf << 20 : 0x0) |
(a & 1 << 4 ? 0xf << 16 : 0x0) |
(a & 1 << 3 ? 0xf << 12 : 0x0) |
(a & 1 << 2 ? 0xf << 8 : 0x0) |
(a & 1 << 1 ? 0xf << 4 : 0x0) |
(a & 1 ? 0xf : 0x0);
}
回答4:
A lookup table, as suggested in the answer by rici will provide the highest performance on most platforms. If you prefer a bit-twiddling approach, the optimal solution will depend on the hardware capabilities of your processor, e.g. how fast are shifts, does it have three-input logic operations (like my GPU), how many integer instructions can it execute in parallel? One solution is to transport each bit to the lsb of its target nibble, then fill in each nibble with its lsb value in a second step (a tip of the hat to chqrlie for suggesting the use of lsb instead of msb):
#include <stdint.h>
uint32_t expand_bits_to_nibbles (uint8_t x)
{
uint32_t r;
/* spread bits to lsb in each nibble */
r = ((((uint32_t)x << (4*0-0)) & (1u << (4*0))) |
(((uint32_t)x << (4*1-1)) & (1u << (4*1))) |
(((uint32_t)x << (4*2-2)) & (1u << (4*2))) |
(((uint32_t)x << (4*3-3)) & (1u << (4*3))) |
(((uint32_t)x << (4*4-4)) & (1u << (4*4))) |
(((uint32_t)x << (4*5-5)) & (1u << (4*5))) |
(((uint32_t)x << (4*6-6)) & (1u << (4*6))) |
(((uint32_t)x << (4*7-7)) & (1u << (4*7))));
/* fill in nibbles */
r = (r << 4) - r;
return r;
}
Some quick experiments with Compiler Explorer show that this leads to a particularly efficient code on PowerPC64, for example.
If the processor has a fast integer multiplier, we could use it to shift multiple bits into place at the same time. Here, we would want to use groups of three source bits to avoid collisions:
#include <stdint.h>
uint32_t expand_bits_to_nibbles_mul (uint8_t x)
{
const uint32_t spread3 = (1u << 6) | (1u << 3) | (1u << 0);
const uint8_t bits_lo3 = (1u << 2) | (1u << 1) | (1u << 0);
const uint8_t bits_md3 = (1u << 5) | (1u << 4) | (1u << 3);
const uint8_t bits_hi2 = (1u << 7) | (1u << 6);
const uint32_t nib_lsb = (1u << 28) | (1u << 24) | (1u << 20) | (1u << 16) |
(1u << 12) | (1u << 8) | (1u << 4) | (1u << 0);
uint32_t r;
/* spread bits to lsb in each nibble */
r = (((uint32_t)(x & bits_lo3) * (spread3 << 0)) +
((uint32_t)(x & bits_md3) * (spread3 << 9)) +
((uint32_t)(x & bits_hi2) * (spread3 << 18))) & nib_lsb;
/* fill in nibbles */
r = (r << 4) - r;
return r;
}
Another variant using integer multiply, which is potentially faster on some platforms, uses an idea from this answer. We use a multiply to spread four bits at a time, such that they land within their target nibble. However, we then have to move the bit within the nibble to the nibble's lsb before we can expand the lsb to cover the nibble. We potentially save a multiplication at the expense of additional housekeeping.
#include <stdint.h>
uint32_t expand_bits_to_nibbles_mul2 (uint8_t x)
{
const uint32_t spread4 = (1u << 12) | (1u << 8) | (1u << 4) | (1u << 0);
const uint32_t extract = (1u << (3*4+3+16)) | (1u << (2*4+2+16)) |
(1u << (1*4+1+16)) | (1u << (0*4+0+16)) |
(1u << (3*4+3+ 0)) | (1u << (2*4+2+ 0)) |
(1u << (1*4+1+ 0)) | (1u << (0*4+0+ 0));
const uint32_t nib_lsb = (1u << 28) | (1u << 24) | (1u << 20) | (1u << 16) |
(1u << 12) | (1u << 8) | (1u << 4) | (1u << 0);
const uint32_t nib_msb = (nib_lsb << 3);
const uint8_t bits_lo4 = (1u << 3) | (1u << 2) | (1u << 1) | (1u << 0);
const uint8_t bits_hi4 = (1u << 7) | (1u << 6) | (1u << 5) | (1u << 4);
uint32_t r;
/* spread bits to their target nibbles */
r = (((uint32_t)(x & bits_lo4) * (spread4 << 0)) +
((uint32_t)(x & bits_hi4) * (spread4 << 12)));
/* extract appropriate bit in each nibble and move it into nibble's lsb */
r = (((r & extract) + (nib_msb - extract)) >> 3) & nib_lsb;
/* fill in each nibble with its lsb */
r = (r << 4) - r;
return r;
}
来源：https://stackoverflow.com/questions/55051490/bit-duplication-from-8-bit-to-32-bit