Bit duplication from 8-bit to 32-bit

Question

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.

Answer 1

It's simple - solve the simplest case, then do more complex ones.

Case 1: Duplicate 1 bit into an 4 bit value (the simplest).

+---+---------+
| 0 | _ _ _ A |
+---+---------+
| 1 | A A A A |
+---+---------+

This can be done as a simple set of shifts:

x = (x << 0) | (x << 1) | (x << 2) | (x << 3);

Or in a less obvious but faster way:

x = (x << 4) - x;

This step will be the last step in all following cases.

Case 2: Duplicate 2 bits into an 8 bit value.

+---+---------+---------+
| 0 | _ _ _ _ | _ _ A B |
+---+---------+---------+
| 1 | _ _ _ A | _ _ _ B |
+---+---------+---------+
| 2 | A A A A | B B B B |
+---+---------+---------+

Case 3: 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 4: 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);
}