I'm looking into why black cursors disappear on a black background, I found this comment in the code: windows/x11drv/mouse.c /* We have to do some magic here, as cursors are not fully * compatible between Windows and X11. Under X11, there * are only 3 possible color cursor: black, white and * masked. So we map the 4th Windows color (invert the * bits on the screen) to black. This require some boolean * arithmetic: * * Windows | X11 * And Xor Result | Bits Mask Result * 0 0 black | 0 1 background * 0 1 white | 1 1 foreground * 1 0 no change | X 0 no change * 1 1 inverted | 0 1 background * * which gives: * Bits = 'And' xor 'Xor' (the X will be 1) * Mask = (not 'And') or 'Xor' * * FIXME: apparently some servers do support 'inverted' color. * I don't know if it's correct per the X spec, but maybe * we ought to take advantage of it. -- AJ */ Simply "OR'ing" the black on black fixes black, but clobbers the white. How do you implement inverted color?(and test the server) I'd like to "take advantage of it" if I can. Or is there a better way to fix this problem? -Jim
