x |= 0xFF << 24
#3

To answer your question: due to operator precedence, first this part is executed:
pawn Код:
0xFF << 24
This shifts 0xFF left by 24 bits. This actually exactly the same as writing 0xFF000000. This value is then OR'd with x. I presume x to be an ARGB color for objects materials, in which case it will make that color completely opaque.
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x |= 0xFF << 24 - by seler - 06.06.2015, 19:44
Re: x |= 0xFF << 24 - by dominik523 - 06.06.2015, 19:56
Re: x |= 0xFF << 24 - by Vince - 06.06.2015, 22:35

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