The Palm OS Reference states "Use WinScreenGetAttribute to retrieve the true hardware dimensions of the display." But as far as I can tell, it doesn't work at all. On the Sony UX40/50 simulator, it returns 480 correctly when asking for winScreenRowBytes, but return 320 when asking for winScreenWidth even if the DIA is collapsed. On the T3 simulator it returns 28 (yes, 28) for winScreenRowBytes and 320/448 for winScreenWidth.
Palm OS seems to point to WinScreenGetAttribute for getting the hardware dimensions of the display, but it doesn't return the correct values if this is true. If you have a landscape device, try it on there. Maybe it'll work on an actual device. Perhaps someone else can shed some light? Oh, if you're just interested in getting the extent of the display not including the DIA area, you can use WinGetDisplayExtent which is very reliable. -- Tim Kostka http://www.nuprograms.com "palmDev" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > i've searched the archive and read the Reference Manuals > and i can not find a way to determine the physical size of the > screen. > > On the Sony UX50 it can be 320x320 or 480x320. I want to > save the bitmap but when the screen is in full landscape calling > BmpGetDimensions() returns 320x320 and not 480x320. > > why isnt there any API to get the physical screen size? > > > > -- For information on using the Palm Developer Forums, or to unsubscribe, please see http://www.palmos.com/dev/support/forums/
