如何在SDL 2.0中指定“宽度”或“点大小”以绘制点、线或矩形

Question

我正在尝试使用SDL 2.0函数 SDL_RenderDrawPoints() 在屏幕上绘制数据点。

在"processing.org"中，我可以使用strokeWeight()来更改我的“点”的大小。如何在SDL 2.0中实现此功能？

- Zhe Hu

Answer 1

SDL本身不支持此功能，需要使用SDL_gfx库。

函数thicklineRGBA可以指定线条宽度。

int thickLineRGBA (SDL_Renderer *rd, Sint16 x1, Sint16 y1, Sint16 x2, Sint16 y2, Uint8 width, Uint8 r, Uint8 g, Uint8 b, Uint8 a)

- Mahesh Bansod

Answer 2

SDL_RenderSetScale

int SDL_RenderSetScale(SDL_Renderer* renderer,
                   float         scaleX,
                   float         scaleY)

注意

在使用渲染器之前，绘图坐标将按照x/y缩放因子进行缩放。这样可以使用单个坐标系统进行分辨率无关的绘图。

如果这导致渲染后端进行缩放或子像素绘制，则将使用适当的质量提示进行处理。为获得最佳效果，请使用整数缩放因子。

引自SDL Wiki

示例

#include <SDL2/SDL.h>

#include <iostream>

int main()
{
    SDL_Renderer* renderer;
    SDL_Window* window;
    SDL_Point points[4];
    SDL_Point  startingPoint;
    startingPoint.x = 50;
    startingPoint.y = 50;
    float scale = 1.0;

    if ( SDL_Init( SDL_INIT_EVERYTHING ) != 0 )
        std::cout << "Failed to init SDL : " << SDL_GetError();

    window = SDL_CreateWindow( "Client", 50, 50, 500, 500, 0 );

    if ( window == nullptr )
        std::cout << "Failed to apply video mode : " << SDL_GetError();

    renderer = SDL_CreateRenderer( window, -1, SDL_RENDERER_ACCELERATED );

    if ( renderer == nullptr )
        std::cout << "Could not create renderer!";

    SDL_RenderSetLogicalSize( renderer, 500, 500 );

    // Clear background
    SDL_SetRenderDrawColor( renderer, 0, 0, 0, 255 );
    SDL_RenderClear( renderer );
    SDL_SetRenderDrawColor( renderer, 255, 255, 255, 255 );

    // Create first 4 points
    points[0].x = startingPoint.x;
    points[0].y = startingPoint.y;

    points[1].x = startingPoint.x + 50;
    points[1].y = startingPoint.y;

    points[2].x = startingPoint.x;
    points[2].y = startingPoint.y + 50;

    points[3].x = startingPoint.x + 50;
    points[3].y = startingPoint.y + 50;

    SDL_RenderDrawPoints( renderer, points, 4 );

    // Create seconds 4 points
    startingPoint.x = 125;
    scale = 2.0;

    points[0].x = startingPoint.x;
    points[0].y = startingPoint.y;

    points[1].x = startingPoint.x + 50;
    points[1].y = startingPoint.y;

    points[2].x = startingPoint.x;
    points[2].y = startingPoint.y + 50;

    points[3].x = startingPoint.x + 50;
    points[3].y = startingPoint.y + 50;

    // Apply scale
    for ( int i = 0; i < 4 ; ++i )
    {
        points[i].x /= scale;
        points[i].y /= scale;
    }

    SDL_RenderSetScale( renderer, scale, scale );
    SDL_RenderDrawPoints( renderer, points, 4 );

    // Create third 4 points
    startingPoint.x = 200;
    scale = 3.0;

    points[0].x = startingPoint.x;
    points[0].y = startingPoint.y;

    points[1].x = startingPoint.x + 50;
    points[1].y = startingPoint.y;

    points[2].x = startingPoint.x;
    points[2].y = startingPoint.y + 50;

    points[3].x = startingPoint.x + 50;
    points[3].y = startingPoint.y + 50;

    // Apply scale
    for ( int i = 0; i < 4 ; ++i )
    {
        points[i].x /= scale;
        points[i].y /= scale;
    }

    SDL_RenderSetScale( renderer, scale, scale );
    SDL_RenderDrawPoints( renderer, points, 4 );

    SDL_RenderPresent( renderer );

    std::cin.ignore();
}

这个例子将绘制一个正方形模式内的三个系列，每个系列包含四个点：

• 1.0 比例尺在 50, 50 到 100, 100
• 2.0 比例尺在 125, 50 到 175, 100
• 3.0 比例尺在 200, 50 到 250, 100

- olevegard

---

这会对渲染器的全部内容进行缩放，因此不仅会影响线条粗细，正如 OP 所请求的那样。 - Timmos

Answer 3

无法在评论中添加，因此以答案的形式呈现。

任何尝试使用@olevegard解决方案的人请注意，这仅缩放SDL_RenderDrawPoints调用 - 例如，对于对角线线条不起作用

您可以将SDL_RenderDrawPoints更改为SDL_RenderDrawLines，并查看结果，如此屏幕截图所示（我已添加了附加级别scale = 6;）

SDL_RenderDrawLines

SDL_RenderDrawPoints

- yatsek