How does multiplying by 66049 duplicate bits?

user_17665807 0
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This is part of a pixel blending operation to enhance precision.

typedef unsigned uint;

uint h(uint x) {
    x &= 0xff;
    return x << 8 | x;
}

uint g(uint x, uint y) {
    return h(x) * h(y) >> 24;
}

I looked at the compiler output, and found a very interesting line.

g:
        movzx   edi, dil
        movzx   esi, sil
        imul    esi, edi
        imul    eax, esi, 66049  # <---
        shr     eax, 24
        ret

This can be decompiled as,

uint g_(uint x, uint y) {
    return (x & 0xff) * (y & 0xff) * 66049 >> 24;
}

I couldn't understand how multiplying by 66049 can produce the desired result. What is the math behind it?

bit-manipulation alphablending x86 assembly c
1 Answers

66049 is 0x10201 hexadecimal. That's the big clue, since:

(x << 8 | x) = (x * 0x101)

(assuming 0 ≤ x ≤ 0xFF), then (x << 8 | x) * (y << 8 | y) is the same as x * y * 0x101 * 0x101.

The 0x101 * 0x101 can be reduced to 0x10201 - there's your magic constant.

So why is x * 0x101 the same as (x << 8 | x)? That should be easy to see with a decimal example. If you multiply any number 0-99 by 101, it duplicates the digits - for example, 32 * 101 = 3232. This is because multiplying by 100 effectively just adds 0s at the end (i.e. 32 * 100 = 3200) and the extra 1 just adds the original value (32 * 101 = 32 * 100 + 32 * 1 = 3200 + 32 = 3232). The same principle applies in hexadecimal (or in binary).

user_3886610
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