按位运算没有任何意义

Question

my current understanding of bitwise operations is that it would push the binary reprisentation of a number a specific amount of times either removing numbers in the process (in case of >> ) or add 0's in the end of a number (in case of << ). so why is it so when i have a int32 storing a hex value of int32_t color = 0xFFFF99FF; (= 1111 1111 1111 1111 1001 1001 1111 1111) int32_t color = 0xFFFF99FF; (= 1111 1111 1111 1111 1001 1001 1111 1111) bitshifting right this int by 24 should give the value FF, because we moved the first two byts by the number of the byts remaining (32 - 8 = 24) but what actually happens is that i end up with the value -1 when i execute my code and 0 in calculator note: bitshifting right by 18 yeilds me the desired result.

am using the SDL2 library and C++

am trying to store colors as their hex values then extract the red,green and blue chanel ignoring the alpha one. the code here is minimized without any unacessary details.

int32_t color; //hex value of the color yellow


//taking input and changing the value of color based on it

if (event->type == SDL_KEYDOWN) {
    switch (event->key.keysym.sym)
    {
        case SDLK_a:
            //SDL_SetRenderDrawColor(renderer, 255, 255, 153, 255); // sand 
            color = 0xFFFF99FF; //hex code for sand color 
            break;
        case SDLK_z:
            //SDL_SetRenderDrawColor(renderer, 0, 0, 255, 255); // water
            color = 0x0000FFFF; //hex code for blue color ...
            break;
        case SDLK_e:
            //SDL_SetRenderDrawColor(renderer, 139, 69, 19, 255); // dirt
            color = 0x8B4513FF;
            break;
        case SDLK_d:
            //SDL_SetRenderDrawColor(renderer, 0, 0, 0, 255); // delete button || air
            color = 0x000000FF;
            break;
        default:
            OutputDebugString("unhandled input.\n");
            break;
    }
}

//checking for mouse input and drawing a pixel with a specific color based on it
if (event-\>button.button == SDL_BUTTON_LEFT)
{
        SDL_SetRenderDrawColor(renderer, color \>\> 24, (color \>\> 16) - (color \>\> 24), (color \>\> 8) - ((color \>\> 16) - (color \>\> 24) + color \>\> 24));
        OutputDebugString(std::to_string(color \>\> 24).c_str());
        SDL_RenderDrawPoint(renderer, mouseX / 4, mouseY / 4);
        SDL_RenderPresent(renderer);
}

Answer 1

int32_t is signed and then >> is the signed shift, keeping the sign bit 31. You should use uint32_t . And it also is more logical to use uint8_t for color components.