本文整理汇总了Java中sun.awt.geom.Curve.pointCrossingsForPath方法的典型用法代码示例。如果您正苦于以下问题：Java Curve.pointCrossingsForPath方法的具体用法？Java Curve.pointCrossingsForPath怎么用？Java Curve.pointCrossingsForPath使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类sun.awt.geom.Curve的用法示例。
在下文中一共展示了Curve.pointCrossingsForPath方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Java代码示例。
示例1: contains
import sun.awt.geom.Curve; //导入方法依赖的package包/类
/**
* Tests if the specified coordinates are inside the closed
* boundary of the specified {@link PathIterator}.
* <p>
* This method provides a basic facility for implementors of
* the {@link Shape} interface to implement support for the
* {@link Shape#contains(double, double)} method.
*
* @param pi the specified {@code PathIterator}
* @param x the specified X coordinate
* @param y the specified Y coordinate
* @return {@code true} if the specified coordinates are inside the
* specified {@code PathIterator}; {@code false} otherwise
* @since 1.6
*/
public static boolean contains(PathIterator pi, double x, double y) {
if (x * 0.0 + y * 0.0 == 0.0) {
/* N * 0.0 is 0.0 only if N is finite.
* Here we know that both x and y are finite.
*/
int mask = (pi.getWindingRule() == WIND_NON_ZERO ? -1 : 1);
int cross = Curve.pointCrossingsForPath(pi, x, y);
return ((cross & mask) != 0);
} else {
/* Either x or y was infinite or NaN.
* A NaN always produces a negative response to any test
* and Infinity values cannot be "inside" any path so
* they should return false as well.
*/
return false;
}
}