java.util.Stack.lastIndexOf()方法的使用及代码示例

x33g5p2x  于2022-01-29 转载在 其他  
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本文整理了Java中java.util.Stack.lastIndexOf()方法的一些代码示例，展示了Stack.lastIndexOf()的具体用法。这些代码示例主要来源于Github/Stackoverflow/Maven等平台，是从一些精选项目中提取出来的代码，具有较强的参考意义，能在一定程度帮忙到你。Stack.lastIndexOf()方法的具体详情如下：
包路径：java.util.Stack
类名称：Stack
方法名：lastIndexOf

Stack.lastIndexOf介绍

暂无

代码示例

代码示例来源：origin: com.mysema.rdf/rdfbean-core

  1. private boolean inNegation() {
  2.   int notIndex = operatorStack.lastIndexOf(Ops.NOT);
  3.   if (notIndex > -1) {
  4.     int existsIndex = operatorStack.lastIndexOf(Ops.EXISTS);
  5.     return notIndex > existsIndex;
  6.   }
  7.   return false;
  8. }

代码示例来源：origin: org.jdom/jdom-legacy

  1. /**
  2.  * Given a prefix, this will return the namespace URI most 
  3.  * rencently (topmost) associated with that prefix.
  4.  *
  5.  * @param prefix <code>String</code> namespace prefix.
  6.  * @return <code>String</code> - the namespace URI for that prefix.
  7.  */
  8. public String getURI(String prefix) {
  9.   int index = prefixes.lastIndexOf(prefix);
  10.   if (index == -1) {
  11.     return null;
  12.   }
  13.   String uri = (String)uris.elementAt(index);
  14.   return uri;       
  15. }

代码示例来源：origin: org.sonatype.maven.archetype/archetype-common

  1. /**
  2.  * Given a prefix, this will return the namespace URI most rencently (topmost) associated with
  3.  * that prefix.
  4.  * 
  5.  * @param prefix <code>String</code> namespace prefix.
  6.  * @return <code>String</code> - the namespace URI for that prefix.
  7.  */
  8. public String getURI(String prefix) {
  9.   int index=prefixes.lastIndexOf(prefix);
  10.   if (index == -1) {
  11.     return null;
  12.   }
  13.   return uris.elementAt(index);
  14. }

代码示例来源：origin: dragome/dragome-sdk

  1. /**
  2.    * Returns the index of the first occurrence of the object.
  3.    *
  4.    * @return the index of the first occurrence of the object
  5.    * @param o
  6.    *            the object to be searched
  7.    */
  8.   public synchronized int search(Object o)
  9.   {
  10.     int index= lastIndexOf(o);
  11.     if (index >= 0)
  12.       return (size() - index);
  13.     return -1;
  14.   }
  15. }

代码示例来源：origin: apache/maven-archetype

  1. /**
  2.  * Given a prefix, this will return the namespace URI most
  3.  * rencently (topmost) associated with that prefix.
  4.  *
  5.  * @param prefix <code>String</code> namespace prefix.
  6.  * @return <code>String</code> - the namespace URI for that prefix.
  7.  */
  8. public String getURI( String prefix )
  9. {
  10.   int index = prefixes.lastIndexOf( prefix );
  11.   if ( index == -1 )
  12.   {
  13.     return null;
  14.   }
  15.   String uri = (String) uris.elementAt( index );
  16.   return uri;
  17. }

代码示例来源：origin: org.apidesign.bck2brwsr/emul

  1. /**
  2.  * Returns the 1-based position where an object is on this stack.
  3.  * If the object <tt>o</tt> occurs as an item in this stack, this
  4.  * method returns the distance from the top of the stack of the
  5.  * occurrence nearest the top of the stack; the topmost item on the
  6.  * stack is considered to be at distance <tt>1</tt>. The <tt>equals</tt>
  7.  * method is used to compare <tt>o</tt> to the
  8.  * items in this stack.
  9.  *
  10.  * @param   o   the desired object.
  11.  * @return  the 1-based position from the top of the stack where
  12.  *          the object is located; the return value <code>-1</code>
  13.  *          indicates that the object is not on the stack.
  14.  */
  15. public synchronized int search(Object o) {
  16.   int i = lastIndexOf(o);
  17.   if (i >= 0) {
  18.     return size() - i;
  19.   }
  20.   return -1;
  21. }

代码示例来源：origin: jtulach/bck2brwsr

  1. /**
  2.  * Returns the 1-based position where an object is on this stack.
  3.  * If the object <tt>o</tt> occurs as an item in this stack, this
  4.  * method returns the distance from the top of the stack of the
  5.  * occurrence nearest the top of the stack; the topmost item on the
  6.  * stack is considered to be at distance <tt>1</tt>. The <tt>equals</tt>
  7.  * method is used to compare <tt>o</tt> to the
  8.  * items in this stack.
  9.  *
  10.  * @param   o   the desired object.
  11.  * @return  the 1-based position from the top of the stack where
  12.  *          the object is located; the return value <code>-1</code>
  13.  *          indicates that the object is not on the stack.
  14.  */
  15. public synchronized int search(Object o) {
  16.   int i = lastIndexOf(o);
  17.   if (i >= 0) {
  18.     return size() - i;
  19.   }
  20.   return -1;
  21. }

代码示例来源：origin: jpox/jpox

  1. /**
  2.  * Method to retrieve the last position of the element.
  3.  * @param element The element
  4.  * @return The last position of this element in the List.
  5.  **/
  6. public int lastIndexOf(Object element)
  7. {
  8.   if (useCache)
  9.   {
  10.     loadFromStore();
  11.   }
  12.   else if (backingStore != null)
  13.   {
  14.     return backingStore.lastIndexOf(ownerSM, element);
  15.   }
  16.   return delegate.lastIndexOf(element);
  17. }

代码示例来源：origin: org.apache.jackrabbit/jackrabbit-core

  1. throws InvalidNodeTypeDefException, RepositoryException {
  2. for (Name nt : supertypes) {
  3.   int pos = inheritanceChain.lastIndexOf(nt);
  4.   if (pos >= 0) {
  5.     StringBuilder buf = new StringBuilder();

代码示例来源：origin: apache/jackrabbit

  1. throws InvalidNodeTypeDefException, RepositoryException {
  2. for (Name nt : supertypes) {
  3.   int pos = inheritanceChain.lastIndexOf(nt);
  4.   if (pos >= 0) {
  5.     StringBuilder buf = new StringBuilder();

代码示例来源：origin: org.apache.jackrabbit/jackrabbit-jcr2spi

  1. for (int i = 0; i < supertypes.length; i++) {
  2.   Name stName = supertypes[i];
  3.   int pos = inheritanceChain.lastIndexOf(stName);
  4.   if (pos >= 0) {
  5.     StringBuffer buf = new StringBuffer();

代码示例来源：origin: org.eclipse.jetty/jetty-start

  1. int idx = seenStack.lastIndexOf(property);
  2. for (int i = idx; i < seenStack.size(); i++)

代码示例来源：origin: apache/jackrabbit

  1. for (int i = 0; i < supertypes.length; i++) {
  2.   Name stName = supertypes[i];
  3.   int pos = inheritanceChain.lastIndexOf(stName);
  4.   if (pos >= 0) {
  5.     StringBuffer buf = new StringBuffer();

代码示例来源：origin: msv/msv

  1. ReferenceExp child = (ReferenceExp)itr.next();
  2. int idx = items.lastIndexOf(child);
  3. if(idx!=-1) {

代码示例来源：origin: com.sun.xml.bind/jaxb1-impl

  1. ReferenceExp child = (ReferenceExp)itr.next();
  2. int idx = items.lastIndexOf(child);
  3. if(idx!=-1) {

代码示例来源：origin: com.sun.xml.bind/jaxb-extra-osgi

  1. ReferenceExp child = (ReferenceExp)itr.next();
  2. int idx = items.lastIndexOf(child);
  3. if(idx!=-1) {

代码示例来源：origin: kohsuke/msv

  1. ReferenceExp child = (ReferenceExp)itr.next();
  2. int idx = items.lastIndexOf(child);
  3. if(idx!=-1) {

代码示例来源：origin: org.apache.jackrabbit/jackrabbit-core

  1. int pos = definingParentNTs.lastIndexOf(nt);
  2. if (pos >= 0) {
  3.   StringBuilder buf = new StringBuilder();

代码示例来源：origin: apache/jackrabbit

  1. int pos = definingParentNTs.lastIndexOf(nt);
  2. if (pos >= 0) {
  3.   StringBuilder buf = new StringBuilder();

代码示例来源：origin: apache/jackrabbit

  1. for (int i = 0; i < childNodeNTs.length; i++) {
  2.   Name nt = childNodeNTs[i];
  3.   int pos = definingParentNTs.lastIndexOf(nt);
  4.   if (pos >= 0) {
  5.     StringBuffer buf = new StringBuffer();

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