java - 处理/线形图-帮助

我昨天张贴了类似的内容，但一无所获。我今天花了几个小时解决问题，但没有任何进展。

我正在使用Processing（语言），并试图实现一种在两点之间画一条线的方法。 （我不想使用库的line（）方法。）

我的lineCreate方法适用于正斜率，但是对于负斜率却失败。您能帮忙找出原因吗？

这是lineCreate（）代码：

void createLine(int x0, int y0, int x1, int y1){
   //...
  // Handle slanted lines...
  double tempDX = x1 - x0;
  double tempDY = y1 - y0;            // Had to create dx and dy as doubles because typecasting dy/dx to a double data type wasn't working.
  double m = (-tempDY / tempDX);      // m = line slope. (Note - The dy value is negative
  int deltaN = (2 * -dx);        // deltaX is the amount to increment d after choosing the next pixel on the line.
  int deltaNE = (2 * (-dy - dx));      // ...where X is the direction moved for that next pixel.
  int deltaE = (2 * -dy);            // deltaX variables are used below to plot line.
  int deltaSE = (2 * (dy + dx));
  int deltaS = (2 * dx);
  int x = x0;
  int y = y0;
  int d = 0;                            // d = Amount d-value changes from pixel to pixel. Depends on slope.
  int region = 0;                  // region = Variable to store slope region. Different regions require different formulas.
 if(m > 1){                            // if-statement: Initializes d, depending on the slope of the line.
      d = -dy - (2 * dx);                  // If slope is 1-Infiniti. -> Use NE/N initialization for d.
      region = 1;
  }
  else if(m == 1)
    region = 2;
  else if(m > 0 && m < 1){
      d = (2 * -dy) - dx;                  // If slope is 0-1 -> Use NE/E initialization for d.
      region = 3;
  }
  else if(m < 0 && m > -1){
      d = (2 * dy) + dx;                  // If slope is 0-(-1) -> Use E/SE initliazation for d.
      region = 4;
  }
  else if(m == -1)
    region = 5;
  else if(m < -1){
      d = dy + (2 * dx);                  // If slope is (-1)-(-Infiniti) -> Use SE/S initialization for d.
      region = 6;
  }
  while(x < x1){                    // Until points are connected...
        if(region == 1){          // If in region one...
              if(d <= 0){                // and d<=0...
              d += deltaNE;            // Add deltaNE to d, and increment x and y.
              x = x + 1;
              y = y - 1;
            }
            else{
              d += deltaN;        // If d > 0 -> Add deltaN, and increment y.
              y = y - 1;
            }
        }
        else if(region == 2){
             x = x + 1;
             y = y - 1;
        }
        else if(region == 3){      // If region two...
                if(d <= 0){
              d += deltaE;
              x = x + 1;
            }
            else{
              d += deltaNE;
              x = x + 1;
              y = y - 1;
            }
        }
        else if(region == 4){        // If region three...
              if(d <= 0){
              d += deltaSE;
              x = x + 1;
              y = y + 1;
            }
            else{
              d += deltaE;
              x = x + 1;
            }
        }
        else if(region == 5){
             x = x + 1;
             y = y + 1;
        }
        else if(region == 6){        // If region four...
              if(d <= 0){
              d += deltaSE;
              x = x + 1;
              y = y + 1;
            }
            else{
              d += deltaS;
              y = y + 1;
            }
          }
        point(x, y);          // Paints new pixel on line going towards (x1,y1).
  }
  return;
}

最佳答案

看看这个page。它用代码示例解释了线描背后的整个理论。

线条绘制有许多已知的算法。阅读有关它们的信息here。