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Question by Chimera3D · Jul 11, 2015 at 01:47 AM · shaderintegeroperatorbitwise

Bitwise & Operator in Shaders?

My question is how can I emulate a bitwise & operator in a shader? I have a line of code that is basically:

int a = i & 255;

(where i can become a relatively large number) in a C# script and I need it to work identically in a shader. How can I emulate this operator in a shader?

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Answer by _Gkxd · Jul 11, 2015 at 06:07 AM

The last time I checked, bitwise operators aren't available in CG. See the description of table 3-2 on this page:

Table 3-2 presents the complete list of operators, along with their precedence, associativity, and usage. Operators marked with a reverse highlight are currently reserved. However, no existing Cg profiles support these reserved operators because current graphics hardware does not support bitwise integer operations.

The bitwise and (&) with 255 is equivalent to modding (%) with 256 though, so you can use that instead.

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avatar image Chimera3D · Jul 11, 2015 at 11:39 AM 0
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Well I did say "emulate" - I knew they weren't available so I was wondering how I could replicate their functionality... thanks!

avatar image Chimera3D · Jul 11, 2015 at 09:43 PM 0
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After quickly testing this I realized that if "a" (in the example in the OP) is negative then modding and the bitwise & are no longer returning the same number... how can I fix that?

avatar image _Gkxd · Jul 12, 2015 at 02:08 PM 1
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Ah, I didn't think that you'd have negatives.

x = ((x % 256) + 256) % 256 would work with negatives. If you know that your number doesn't go below -256, then you can use x = (x + 256) % 256 ins$$anonymous$$d.

You might also be able to make the number unsigned... it seems that they exist in Cg: http://http.developer.nvidia.com/Cg/Cg_language.html

Actually, now that I think about it, you can probably just use unsigned chars and completely ignore bitwise operators/modding, since integer overflow for unsigned chars would keep the value between 0 and 255.

avatar image Chimera3D · Jul 12, 2015 at 09:57 PM 0
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Thanks again!

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