C++ 如何在SDL2中的不透明矩形上绘制透明或半透明矩形？

下面是我绘制

SDL_Rect

对象

Rect

和

rect2

的代码：

#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;
}

#包括
#包括
int main（）{
SDL_Window*Window=SDL_CreateWindow（“测试”，SDL_WINDOWPOS_居中，SDL_WINDOWPOS_居中，640480，SDL_Window_显示）；
如果（窗口==0）
std：：cout您只需打电话

SDL\u SetRenderDrawBlendMode（渲染器，SDL\u BLENDMODE\u混合）；


在创建渲染器以启用与RenderDraw函数的alpha混合后。（
可能重复的