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Jody
Joined: 08 Sep 2006 Posts: 182
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T6963C image showing |
Posted: Fri Sep 11, 2015 6:25 am |
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Hello,
I am using the T6963C driver found in the code library.. works fine.
However. When I do:
Code: |
glcd_image8(&bt_logo[0],0,0,8,11);
glcd_image8(&bt_logo[0],5,12,8,11);
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both the pictures are shown at x=0.
When I do:
Code: |
glcd_image8(&bt_logo[0],0,0,8,11);
glcd_image8(&bt_logo[0],13,12,8,11);
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the first image shows up at x=0 and the second at x=8.
I can't get the image to show up at the location I want.. It goes in step off 8 bit..
But I can't find the spot where it goes wrong.....
Anybody an idea?
Regards,
Jody |
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PCM programmer
Joined: 06 Sep 2003 Posts: 21708
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Posted: Fri Sep 11, 2015 10:21 am |
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A search finds three T6963C drivers in the code library.
Post a link to the one you are using. |
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Jody
Joined: 08 Sep 2006 Posts: 182
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PCM programmer
Joined: 06 Sep 2003 Posts: 21708
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Posted: Sun Sep 13, 2015 2:47 am |
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It appears that this routine is intended to place images on 8-pixel boundaries.
That's because it calls the glcd_pixel8() function, and if you look at that
function, it divides the x parameter by 8. In other words, it throws away
any resolution finer than 8 bits. Any incoming x parameter from 0 to 7
will be truncated to 0. Any incoming x parameter from 8 to 15 will be
truncated to 8. This is what you are seeing.
Code: | // glcd_pixel8(x,y,px8) sets 8 pixels in line.
void glcd_pixel8(unsigned int8 x, unsigned int8 y, int8 pixel8){
unsigned int8 x_H;
x_H = (x / 8); // *** x parameter is divided by 8 ***
glcd_gotoxy(x_H+1,y,0);
glcd_WriteCmd1(pixel8,0xc0);
} |
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Jody
Joined: 08 Sep 2006 Posts: 182
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Posted: Sun Sep 13, 2015 3:33 am |
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Thanks....
I have to rewrite mine one function for the thing I want...
I will dive in to it... post it if I get it working..
Thanks for the help!! |
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