Convert 0x1234 to 0x11223344

Question

How do I expand the hexadecimal number 0x1234 to 0x11223344 in a high-performance way?

unsigned int c = 0x1234, b;
b = (c & 0xff) << 4 | c & 0xf |          


        

Answer 1

unsigned long transform(unsigned long n)
{

    /* n: 00AR
     *    00GB
     */
    n = ((n & 0xff00) << 8) | (n & 0x00ff);

    /* n: 0AR0
     *    0GB0
     */
    n <<= 4;

    /* n: AAR0
     *    GGB0
     */
    n |= (n & 0x0f000f00L) << 4;

    /* n: AARR
     *    GGBB
     */
    n |= (n & 0x00f000f0L) >> 4;

    return n;
}

The alpha and red components are shifted into the higher 2 bytes where they belong, and the result is then shifted left by 4 bits, resulting in every component being exactly where it needs to be.

With a form of 0AR0 0GB0, a bit mask and left-shift combination is OR'ed with the current value. This copies the A and G components to the position just left of them. The same thing is done for the R and B components, except in the opposite direction.