[SDL] Re: SDL Surfaces and OpenGL

[Richard Schreyer]

If you can set up an SDL_Surface to use 24 bit pixels (ie, no alpha  
data), you can change the glTexImage2D Image format argument from  
GL_RGBA to GL_RGB.

Richard Schreyer

On Jul 15, 2005, at 2:12 AM, Chris E. wrote:

> Stephen Sweeney wrote:
>
>> Hi Kids :)
>>
>> Hope you're all having a lush summer! :)
>>
>> Right... um... I'm making SDL Surfaces and turning them in to OpenGL
>> textures... but all my SDL Surfaces has alpha set on them. So when  
>> I'm
>> drawing on my created surfaces they are always see through... I  
>> wish them to
>> be solid. My code for creating the surface is the usual example  
>> stuff (see
>> all code below).
>>
>> Any help..?
>>
>> Steve
>>
>> SDL_Surface *Graphics::createSurface(int width, int height)
>> {
>>     SDL_Surface *surface;
>>     Uint32 rmask, gmask, bmask, amask;
>>
>>     /* SDL interprets each pixel as a 32-bit number, so our masks  
>> must depend
>>     on the endianness (byte order) of the machine */
>>     #if (SDL_BYTEORDER == SDL_BIG_ENDIAN)
>>         rmask = 0xff000000;
>>         gmask = 0x00ff0000;
>>         bmask = 0x0000ff00;
>>         amask = 0x000000ff;
>>     #else
>>         rmask = 0x000000ff;
>>         gmask = 0x0000ff00;
>>         bmask = 0x00ff0000;
>>         amask = 0xff000000;
>>     #endif
>>
>>     surface = SDL_CreateRGBSurface(SDL_SWSURFACE, width, height,  
>> 32, rmask,
>> gmask, bmask, amask);
>>
>
> Because you are passing a mask for alpha, SDL is thinking that you  
> want to use it.
> To get rid of the alpha, pass 0 to SDL_CreateRGBSurface in place of  
> amask.
> You can also get rid of the declaration of amask, and the lines  
> that assign it a value.
>
> I'm not sure how OpenGL would handle that - it might still  
> interpret the "missing" bits
> as alpha, and use transparency. A safer solution, given that  
> possibility, since OpenGL
> might well require the alpha channel to be there, would be to make  
> sure that you set
> the alpha value to opaque (255 I think) for every pixel (not just  
> the surface alpha
> value), as the transparency might just be because of the bits  
> defaulting to 0, which is
> completely transparent.
>
> Chris E.
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