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我正在尝试在 java 程序中读取 Path2D.Double 的段，而 Eclips 不断告诉我：

PathIterator 类型中的方法 currentSegment(float[]) 不适用于参数 (Double[])

但是在接口 PathIterator 中，有 2 种方法似乎让 Eclipse（或我）感到困惑

/**
 * Returns the coordinates and type of the current path segment in
 * the iteration.
 * The return value is the path-segment type:
 * SEG_MOVETO, SEG_LINETO, SEG_QUADTO, SEG_CUBICTO, or SEG_CLOSE.
 * A float array of length 6 must be passed in and can be used to
 * store the coordinates of the point(s).
 * Each point is stored as a pair of float x,y coordinates.
 * SEG_MOVETO and SEG_LINETO types returns one point,
 * SEG_QUADTO returns two points,
 * SEG_CUBICTO returns 3 points
 * and SEG_CLOSE does not return any points.
 * @param coords an array that holds the data returned from
 * this method
 * @return the path-segment type of the current path segment.
 * @see #SEG_MOVETO
 * @see #SEG_LINETO
 * @see #SEG_QUADTO
 * @see #SEG_CUBICTO
 * @see #SEG_CLOSE
 */
public int currentSegment(float[] coords);

/**
 * Returns the coordinates and type of the current path segment in
 * the iteration.
 * The return value is the path-segment type:
 * SEG_MOVETO, SEG_LINETO, SEG_QUADTO, SEG_CUBICTO, or SEG_CLOSE.
 * A double array of length 6 must be passed in and can be used to
 * store the coordinates of the point(s).
 * Each point is stored as a pair of double x,y coordinates.
 * SEG_MOVETO and SEG_LINETO types returns one point,
 * SEG_QUADTO returns two points,
 * SEG_CUBICTO returns 3 points
 * and SEG_CLOSE does not return any points.
 * @param coords an array that holds the data returned from
 * this method
 * @return the path-segment type of the current path segment.
 * @see #SEG_MOVETO
 * @see #SEG_LINETO
 * @see #SEG_QUADTO
 * @see #SEG_CUBICTO
 * @see #SEG_CLOSE
 */
public int currentSegment(double[] coords);

我自己的代码如下：

PathIterator it= ((Path2D.Double) p3d).getPathIterator(null, 2d);
  Double x, y;
  while (!it.isDone()) {
     Double coords[]= new Double[6];
     System.out.println("1");
     int art= it.currentSegment(coords);
     System.out.println("2");
     switch (art) {
     case PathIterator.SEG_MOVETO:
        movetoXY(coords[0], coords[1]);
        x= coords[0];
        y= coords[1];
        break;
     case PathIterator.SEG_LINETO:
        linetoXY(coords[0], coords[1]);
        break;
     case PathIterator.SEG_CLOSE:
        linetoXY(x, y);
        break;
     default:
        System.out.println("unbekanntes Segment " + art);
        break;
     }
     it.next();
  }

线与

     int art= it.currentSegment(coords);

标记为红色，工具提示显示：

PathIterator 类型中的方法 currentSegment(float[]) 不适用于参数 (Double[])

有一个带有 float[] 的方法和另一个带有 double[] 但名称相同的方法我在这里想念什么？

4

1 回答 1

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Double[] 和 double[] 不是同一类型，因此您需要转换这些类型...

Java8 Streams 可以以一种语法上非常清晰的方式做到这一点：

只有一件重要的事情：Double 是一个对象，所以 Double[] 可以保存一个空值……你需要检查一下！

Double[] d = { 1.0, 2.0, 3.0, 4.0, 5.0, null };
double[] dArray = Arrays.stream(d).mapToDouble(x -> x == null ? 0.0 : x.doubleValue()).toArray();
System.out.println(Arrays.toString(dArray));

如果您 100% 确定您的 Double[] 没有空值，那么只需执行以下操作：

double[] dArray = Arrays.stream(d).mapToDouble(Double::doubleValue).toArray();
于 2017-05-10T15:19:31.060 回答