SDL_DisplayFormat works, but not SDL_DisplayFormatAlpha
- by Bounderby
The following code is intended to display a green square on a black background. It executes, but the green square does not show up. However, if I change SDL_DisplayFormatAlpha to SDL_DisplayFormat the square is rendered correctly.
So what don't I understand? It seems to me that I am creating *surface with an alpha mask and I am using SDL_MapRGBA to map my green color, so it would be consistent to use SDL_DisplayFormatAlpha as well.
(I removed error-checking for clarity, but none of the SDL API calls fail in this example.)
#include <SDL.h>
int main(int argc, const char *argv[])
{
SDL_Init( SDL_INIT_EVERYTHING );
SDL_Surface *screen = SDL_SetVideoMode(
640, 480, 32, SDL_HWSURFACE | SDL_DOUBLEBUF
);
SDL_Surface *temp = SDL_CreateRGBSurface(
SDL_HWSURFACE, 100, 100, 32, 0, 0, 0,
( SDL_BYTEORDER == SDL_BIG_ENDIAN ? 0x000000ff : 0xff000000 )
);
SDL_Surface *surface = SDL_DisplayFormatAlpha( temp );
SDL_FreeSurface( temp );
SDL_FillRect(
surface, &surface->clip_rect, SDL_MapRGBA(
screen->format, 0x00, 0xff, 0x00, 0xff
)
);
SDL_Rect r;
r.x = 50;
r.y = 50;
SDL_BlitSurface( surface, NULL, screen, &r );
SDL_Flip( screen );
SDL_Delay( 1000 );
SDL_Quit();
return 0;
}
