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Java CodePointIterator.next方法代码示例

本文整理汇总了Java中sun.text.CodePointIterator.next方法的典型用法代码示例。如果您正苦于以下问题:Java CodePointIterator.next方法的具体用法?Java CodePointIterator.next怎么用?Java CodePointIterator.next使用的例子?那么恭喜您, 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sun.text.CodePointIterator的用法示例。


在下文中一共展示了CodePointIterator.next方法的2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: nextFontRunIndex

import sun.text.CodePointIterator; //导入方法依赖的package包/类
/**
 * Determines the font index for the code point at the current position in the
 * iterator, then advances the iterator to the first code point that has
 * a different index or until the iterator is DONE, and returns the font index.
 * @param iter a code point iterator, this will be advanced past any code
 *             points that have the same font index
 * @return the font index for the initial code point found, or 1 if the iterator
 * was empty.
 */
public int nextFontRunIndex(CodePointIterator iter) {
    int cp = iter.next();
    int fontIndex = 1;
    if (cp != CodePointIterator.DONE) {
        fontIndex = getFontIndex(cp);

        while ((cp = iter.next()) != CodePointIterator.DONE) {
            if (getFontIndex(cp) != fontIndex) {
                iter.prev();
                break;
            }
        }
    }
    return fontIndex;
}
 
开发者ID:SunburstApps,项目名称:OpenJSharp,代码行数:25,代码来源:FontResolver.java

示例2: requiresBidi

import sun.text.CodePointIterator; //导入方法依赖的package包/类
/**
    * Return true if the specified text requires bidi analysis.  If this returns false,
    * the text will display left-to-right.  Clients can then avoid constructing a Bidi object.
    * Text in the Arabic Presentation Forms area of Unicode is presumed to already be shaped
    * and ordered for display, and so will not cause this function to return true.
    *
    * @param text the text containing the characters to test
    * @param start the start of the range of characters to test
    * @param limit the limit of the range of characters to test
    * @return true if the range of characters requires bidi analysis
    */
   public static boolean requiresBidi(char[] text, int start, int limit) {
CodePointIterator cpi = CodePointIterator.create(text, start, limit);
for (int cp = cpi.next(); cp != CodePointIterator.DONE; cp = cpi.next()) {
    if (cp > 0x0590) {
	int dc = nativeGetDirectionCode(cp);
	if ((RMASK & (1 << dc)) != 0) {
	    return true;
	}
    }
}
return false;
   }
 
开发者ID:jgaltidor,项目名称:VarJ,代码行数:24,代码来源:Bidi.java


注:本文中的sun.text.CodePointIterator.next方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。