`James Simmons wrote:> Lets look at the new code that I have done with your above parameters.>>        for (i = image->height; i--; ) {>                shift = val = 0;>                n = image->width;>                dst = (u32 __iomem *) dst1;>> 		while (n--) {> 			if (!s) { src++; s = 32; }> 			s -= 1;> 			color = (swapb32p(src) & (1 << s)) ? 1 : 0;> 			val |= color << shift;>> 		       /* Did the bitshift spill bits to the next long? */>                        if (shift >= 31) {>                                FB_WRITEL(val, dst++);>                                val = (shift == 31) ? 0 :(color >> (32 - shift));>                        }>                        shift += 1;>                        shift &= (32 - 1);>                }>>                [ ...]>> with 's' taking values from 31 to 0, and 'shift' taking values from 0 to> 31. In the case of bits_per_pixel = 1 we have>> s -= 1;> color = (swapb32p(src) & (1 << s)) ? 1 : 0;> val |= color << shift;I suppose here that you meant 'swab32p' instead of 'swapb32p'. I can'tfind any definition of 'swapb32p' and in your last patch you sent youis using 'swab32p' which converts a 32-bits little endian word into a32-bits big endian one.>> which reduces to val = color;>> I'm I seeing it wrong?Well, I would say yes you are. If src = { 0xa3, 0x30, 0xef, 0x72 ...}	swab32p(src) -> 0x72ef30a3 -> 01110010 11101111 00110000 10100011during loop #1 (s=31, shift=0):	color = 0x72ef30a3 & (1<<31) ? 1 : 0;	color is 0	val |= 0 << 0;				val is 0during loop #2 (s=30, shift=1):	color = 0x72ef30a3 & (1<<30) ? 1 : 0;	color is 1	val |= 1 << 1;				val is 2	during loop #3 (s=29, shift=2):	color = 0x72ef30a3 & (1<<29) ? 1 : 0;	color is 1	val |= 1 << 2;				val is 6during loop #4 (s=28, shift=3):	color = 0x72ef30a3 & (1<<28) ? 1 : 0;	color is 1	val |= 1 << 3;				val is 0xeduring loop #5 (s=27, shift=4):	color = 0x72ef30a3 & (1<<27) ? 1 : 0;	color is 0	val |= 0 << 4;				val is 0xeduring loop #6 (s=26, shift=5):	color = 0x72ef30a3 & (1<<26) ? 1 : 0;	color is 0	val |= 0 << 5;				val is 0xeduring loop #7 (s=25, shift=6):	color = 0x72ef30a3 & (1<<25) ? 1 : 0;	color is 1	val |= 1 << 6;				val is 0x4eduring loop #8 (s=24, shift=7):	color = 0x72ef30a3 & (1<<24) ? 1 : 0;	color is 0	val |= 0 << 7;				val is 0x4eand so on...Finally val -> 11000101 00001100 11110111 01001110 -> 0xc50cf74eand FB_WRITEL(val, dst++) will write { 0x4e, 0xf7, 0x0c, 0xc5} intothe frame buffer.Am I seeing it wrong ?> BTW what is your visual?>FYI, I give you all screen info, maybe something is miss-initialized...static struct fb_fix_screeninfo t6963c_fb_fix __initdata = {	.id		= DRIVER_NAME,	.type		= FB_TYPE_PACKED_PIXELS,	.visual		= FB_VISUAL_MONO01,	.accel		= FB_ACCEL_NONE,};static struct fb_var_screeninfo t6963c_fb_var __initdata = {	.bits_per_pixel	= 1,	.red		= {0, 1, 0},	.green		= {0, 1, 0},	.blue		= {0, 1, 0},	.transp		= {0, 0, 0},	.activate	= FB_ACTIVATE_NOW,	.height		= -1,		/* height of picture in mm */	.width		= -1,		/* width of picture in mm */	.accel_flags	= 0,	.vmode		= FB_VMODE_NONINTERLACED,};		Franck-To unsubscribe from this list: send the line "unsubscribe linux-kernel" inthe body of a message to majordomo@vger.kernel.orgMore majordomo info at  http://vger.kernel.org/majordomo-info.htmlPlease read the FAQ at  http://www.tux.org/lkml/`