glDrawPixels Access violation reading location

Question

I am trying to render volumetric data using OpenGL function glDrawPixels()

#define SIZE    480
unsigned int    rgbPixels[SIZE][SIZE]
...
glDrawPixels(SIZE, SIZE, GL_RGB, GL_UNSIGNED_INT, rgbPixels);

If I do sizeof(rgbpixels) it returns 921600 which is what 480*480*4.

I have tried GL_RGBA but does no good. Finally I got something on the screen without getting access violation using glDrawPixels(SIZE, SIZE*0.81, GL_RGB, GL_UNSIGNED_INT, rgbPixels);

But it is not as per my expected output. So can anyone help me in this case?

Answer 1

You are telling glDrawPixels that your data is SIZExSIZE pixels and each pixel has 3 components ( GL_RGB ) of 32-bit unsigned integer type each ( GL_UNSIGNED_INT ). If this is not the case for rgbPixels (which probably isn't when you say it's just SIZE*SIZE*4 instead of the required SIZE*SIZE*3*4 ), then this will likely result in an access violation or other undefined behaviour.

When you say your rgbPixels is of size SIZE*SIZE*4 , then I guess each pixel contains 4-bytes, with each individual byte being one color component. If this is the case you need

glDrawPixels(SIZE, SIZE, GL_RGBA, GL_UNSIGNED_BYTE, rgbPixels);

Just always keep in mind that all the parameters of glDrawPixels tell the OpenGL how much memory the pointer argument ( rgbPixels ) uses and in which format it is layed out. So those of course have to match the actual memory that rgbPixels contains (or points to).

Answer 2

As given the OpenGL will expect 3 unsigned integers per pixel! You probably want to use GL_UNSIGNED_INT_8_8_8_8 instead as the data type.