Bit hack: Expanding bits

Question

I am trying to convert a uint16_t input to a uint32_t bit mask. One bit in the input toggles two bits in the output bit mask. Here is an example conver

Answer 1

The easiest way to map a 4-bit input to an 8-bit output is with a 16 entry table. So then it's just a matter of extracting 4 bits at a time from the uint16_t, doing a table lookup, and inserting the 8-bit value into the output.

uint32_t expandBits( uint16_t input )
{
    uint32_t table[16] = {
        0x00, 0x03, 0x0c, 0x0f,
        0x30, 0x33, 0x3c, 0x3f,
        0xc0, 0xc3, 0xcc, 0xcf,
        0xf0, 0xf3, 0xfc, 0xff
    };

    uint32_t output;
    output  = table[(input >> 12) & 0xf] << 24;
    output |= table[(input >>  8) & 0xf] << 16;
    output |= table[(input >>  4) & 0xf] <<  8;
    output |= table[ input        & 0xf];
    return output;
}

This provides a decent compromise between performance and readability. It doesn't have quite the performance of cmaster's over-the-top lookup solution, but it's certainly more understandable than thndrwrks' magical mystery solution. As such, it provides a technique that can be applied to a much larger variety of problems, i.e. use a small lookup table to solve a larger problem.

Answer 2

Try this, where input16 is the uint16_t input mask:

uint32_t input32 = (uint32_t) input16;
uint32_t result = 0;
uint32_t i;
for(i=0; i<16; i++)
{
    uint32_t bit_at_i = (input32 & (((uint32_t)1) << i)) >> i;
    result |= ((bit_at_i << (i*2)) | (bit_at_i << ((i*2)+1)));
}
// result is now the 32 bit expanded mask