如何在 SDL2 中的不透明矩形之上绘制透明或半透明矩形？

Question

这是我绘制SDL_Rect对象的代码rect和rect2：

#include <iostream>
#include <SDL2/SDL.h>

int main(){
    SDL_Window *window=     SDL_CreateWindow("Test", SDL_WINDOWPOS_CENTERED, SDL_WINDOWPOS_CENTERED, 640, 480, SDL_WINDOW_SHOWN);
    if(window == 0)
        std::cout << SDL_GetError() << std::endl;

    SDL_Renderer *renderer= SDL_CreateRenderer(window, -1, SDL_RENDERER_ACCELERATED | SDL_RENDERER_PRESENTVSYNC);
    if(renderer == NULL)
        std::cout << SDL_GetError() << std::endl;

    SDL_Rect rect;
    rect.x=     100;
    rect.y=     100;
    rect.w=     100;
    rect.h=     100;

    SDL_Rect rect2;
    rect2.x=    150;
    rect2.y=    150;
    rect2.w=    100;
    rect2.h=    100;

    while(true){
        Uint8 r,g,b,a;

        r=  32;
        g=  32;
        b=  32;
        a=  255;
        if(SDL_SetRenderDrawColor(renderer, r, g, b, a) == -1)
            std::cout << SDL_GetError() << std::endl;
        if(SDL_RenderClear(renderer) == -1)
            std::cout << SDL_GetError() << std::endl;

        r=  255;
        g=  255;
        b=  255;
        a=  255;
        if(SDL_SetRenderDrawColor(renderer, r, g, b, a) == -1)
            std::cout << SDL_GetError() << std::endl;
        if(SDL_RenderFillRect(renderer, &rect) == -1)
            std::cout << SDL_GetError() << std::endl;

        r=  100;
        g=  100;
        b=  100;
        a=  0;
        if(SDL_SetRenderDrawColor(renderer, r, g, b, a) == -1)
            std::cout << SDL_GetError() << std::endl;
        if(SDL_RenderFillRect(renderer, &rect2) == -1)
            std::cout << SDL_GetError() << std::endl;

        SDL_RenderPresent(renderer);
    }

    return 0;
}

虽然我能画rect到SDL_Renderer对象renderer就好了，如果我添加rect2它始终是不透明的，阻断观点rect。即使我将rect2的 alpha设置为 0，它仍然显示为不透明，挡住rect视线。

我该如何解决这个问题，使之rect2更加透明？

Answer 1

您需要做的就是
SDL_SetRenderDrawBlendMode(renderer, SDL_BLENDMODE_BLEND);
在创建渲染器后调用以启用与 RenderDraw 函数的 alpha 混合。（SDL 文档）