fbdev: why is this necessary?

Lee Holeva

Using the ARBG888 format, I need to use the following bit of code to get the pixels into the frame buffer memory:

pmem = addr;
for(i=0;i<N;i++){
    offset=0;
    for(j=0;j<M;j++){
      a4bpix.byte[0] = 0x00;
      a4bpix.byte[1] = ibuf[i][j];
      a4bpix.byte[2] = ibuf[i][j];
      a4bpix.byte[3] = ibuf[i][j];
      memcpy(pmem+offset, a4bpix.byte, 4);
      offset+=4;
    }
    pmem+=((M*32)>>3);
}

question: Why do I need to multiply by 32 and shift right by 3 bits, multiply by 4 all together) the pointer after each line? If I do not do this I get a segmentation fault. What is going on in the hardware that makes the pointer adjustment necessary?

Lee Holeva

over 14 years ago

0 yihe over 14 years ago

TI__Guru 66120 points

Holeva,

I assume that pmem is a u8 type pointer.

for every N you need adjust pmem to move the next line. that is right procedure. you can do pmem += M*4;

if you do not change pmem for every i(N), you just copy the different data to the first line again and again, but I did not see how you can get a segmentation fault.

In order to assit you, please provide your full souce codes

0 Lee Holeva over 14 years ago in reply to yihe

Genius 4615 points

Actually, I do not get a segmentation fault using the above code. If I simply increment pmem by M, the number of bytes per line, I see four liitle copies of the image displayed. If on the other hand, I increment pmem by M*32, I get a segmentation fault. My question is why is the multiplcation of the increment of pmem by 4 required? What is going on in the hardware that makes this necessay?

0 Steve Clynes over 14 years ago in reply to Lee Holeva

TI__Mastermind 43875 points

Lee,

You certainly don't want to increment pmem by M*32 since this will cause a buffer over flow. You should increment by M*4 (which is what your example code does) since there are 4 bytes per pixel. If you try to access non word aligned addresses then you might also get an error.

How big are M & N?

At what point to you get the seg fault? (i.e. can you print M & N during the run to see if it is possibly a buffer over run?)

I would update pmem inside the inner loop by adding 4 every loop rather than updating offset unless you plan on doing some line based optimizations later.

BR,

Steve

0 Lee Holeva over 14 years ago in reply to Steve Clynes

Genius 4615 points

M=480 N=720

pmem is a pointer to unsigned char and ibuf is a 2D array of unsigned char. Incrementing pmem by 4 while writing each line makes sense, but the need to multiply pmem by 4 at the end of each line is mysterious. It looks like write a line of pixels to the framebuffer and then skip three lines before writing the next line. What is going on in the video back-end hardware that requires this? A possible clue is that if I do not skip three lines I get not a single image, but four duplicate little images.

Lee Holeva

0 Norman Wong over 14 years ago in reply to Lee Holeva

Guru 26430 points

Four bytes are moved each loop of M. Makes sense to increment pmem by M*4 bytes after M loops as pmem is NOT incremented in the M loop itself. The multiply and shift seem a bit redundant. Shifting left by 5 and then right by 3 is the same as shifting left by 2. Unless shifting left by 5 will move significant bits off the end. I would suspect most compilers would just optimize that out to a single shift. Implied in your code fragment is a grayscale to RGB888 conversion where each pixel is big endian and padded to 32 bits. Something like this.

#define N 720
#define M 480
void grayscale_to_aligned_rgb888
( const unsigned char grayscale[N][M],
unsigned char rgb888[N*M*4]
)
{
register unsigned char       *pdst = rgb888;
register const unsigned char *psrc = grayscale;
register unsigned char        v;
for(i=0;i<N;i++)
{
    for(j=0;j<M;j++)
    { v      = *psrc++;
      *pdst++ = 0x00; /* Pad, MSB*/
      *pdst++ = v;    /* Red */
      *pdst++ = v;    /* Green */
      *pdst++ = v;    /* Blue, LSB */
    }
}
}
Is your intent something else entirely?

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fbdev: why is this necessary?