Re: Get RGBA image
Suhail Doshi <[email protected]> Mon, 5 Aug 2019 18:07:04 -0700
| Newsgroups | gmane.comp.freedesktop.xcb |
|---|---|
| Message-ID | <CAKDSyZ6QgibkaqhqgkrrvC3m+opL=VRbdcJP+O-NFuQQJ_cp1w@mail.gmail.com> |
Is there a way to just get 8-bits per channel (3 or 4 channel)? I am using an encoder that accepts 8-bit packed RGBA. Ideally, I don’t have to do bit manipulation on my own because that will slow things down. On Mon, Aug 5, 2019 at 5:56 PM Alan Coopersmith <[email protected]> wrote: > On 8/5/19 5:51 PM, Suhail Doshi wrote: > > Hi there, > > > > I am trying out this > > code: https://gist.github.com/Suhail/3ae62751857ff034a027adcacd350b69 > > > > One thing I noticed is that when I print the xcb image struct info I > don't get > > the image size I was expecting: > > > > xcb_image_print() Printing a (1024,768) xcb_image_t of 1572864 bytes, > depth: 16, > > bpp: 16 > > > > Do you know why ximg->size wouldn't be 1024*768*3 (3 channel: RGB)? Or 4 > channel? > > > > 1572864 seems like an odd size. > > It's the size I'd expect - 1024 * 768 * 2 (where 2 is 16 bits-per-pixel > divided > by 8 bits per byte). You might have 4 bits each of RGBA or 5 bits each of > RGB > plus a pad bit, or some other combination of channels, in those 16 bits. > > -- > -Alan Coopersmith- [email protected] > Oracle Solaris Engineering - https://blogs.oracle.com/alanc > -- Suhail -- Founder _______________________________________________ Xcb mailing list [email protected] https://lists.freedesktop.org/mailman/listinfo/xcb