Re: [Linux-fbdev-devel] fbmem: is bootup logo broken for monochrome LCD ?

[Franck Bui-Huu]

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't
find any definition of 'swapb32p' and in your last patch you sent you
is using 'swab32p' which converts a 32-bits little endian word into a
32-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 10100011


during loop #1 (s=31, shift=0):

	color = 0x72ef30a3 & (1<<31) ? 1 : 0;	color is 0
	val |= 0 << 0;				val is 0

during 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 6

during loop #4 (s=28, shift=3):

	color = 0x72ef30a3 & (1<<28) ? 1 : 0;	color is 1
	val |= 1 << 3;				val is 0xe

during loop #5 (s=27, shift=4):

	color = 0x72ef30a3 & (1<<27) ? 1 : 0;	color is 0
	val |= 0 << 4;				val is 0xe

during loop #6 (s=26, shift=5):

	color = 0x72ef30a3 & (1<<26) ? 1 : 0;	color is 0
	val |= 0 << 5;				val is 0xe

during loop #7 (s=25, shift=6):

	color = 0x72ef30a3 & (1<<25) ? 1 : 0;	color is 1
	val |= 1 << 6;				val is 0x4e

during loop #8 (s=24, shift=7):

	color = 0x72ef30a3 & (1<<24) ? 1 : 0;	color is 0
	val |= 0 << 7;				val is 0x4e

and so on...

Finally val -> 11000101 00001100 11110111 01001110 -> 0xc50cf74e

and FB_WRITEL(val, dst++) will write { 0x4e, 0xf7, 0x0c, 0xc5} into
the 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
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