uFlags and

[nwscomps]

if I understand well the uFlags parameter always assume that each uflag bit holds 1 byte. In your documentation you make an example of processing R,G,B channels using 1 | 2 | 4 flags. Is it me or I cannot understand why? Let's take a 24 bits image 3 channels (R,G,B) 1 byte per channel. I would expect the flags to be 0| 1| 2 can you explain why is it not like this. I can't seem to graps the logic behind.
You write in your documentation also: "If bit N is set, channel N is processed. If bit N is not set, channel N is skipped"
Thanks.

[Admin]

"each uflag bit holds 1 byte"

Each uFlags bit is one bit. There are 32 bits in the four bytes of uFlags' UINT32.

If you are setting bits 0, 1, 2, then the value in uFlags is 7, because:

00000000000000000000000000000001 = 1
00000000000000000000000000000010 = 2
00000000000000000000000000000100 = 4

And 1 | 2 | 4 = 7
"|" is binary OR.


You write in your documentation also: "If bit N is set, channel N is processed. If bit N is not set, channel N is skipped"
Correct.
Ex. If you have a 3 channel image (R,G,B), you can skip channel 0 by setting only bits 1 and 2 (2 | 4 = 6). Or you can process just the Red channel (channel 0) by setting just bit 0: which is a value of1.

[nwscomps]

Thanks a lot I think I got it, but since I am not a binary lover :O (please don't jump on me just to make sure would you be so kind to describe the setting of uflags for the following case:

Process RGBA (RGB+4th byte=alpha channel) 8 bits per channel
Process RGBA (RGB+4th byte=alpha channel) 16 bits per channel

thanks


[QUOTE=Admin;11256]"each uflag bit holds 1 byte"

Each uFlags bit is one bit. There are 32 bits in the four bytes of uFlags' UINT32.

If you are setting bits 0, 1, 2, then the value in uFlags is 7, because:

00000000000000000000000000000001 = 1
00000000000000000000000000000010 = 2
00000000000000000000000000000100 = 4

And 1 | 2 | 4 = 7
"|" is binary OR.

[Admin]

00000000000000000000000000000001 = 1 
00000000000000000000000000000010 = 2 
00000000000000000000000000000100 = 4
00000000000000000000000000001000 = 8

so OR them together (which, since none of the bits overlap, is equivalent to addition), all four channels = 15 (1 | 2 | 4 | 8)

00000000000000000000000000001111 = 15