Java Eclipse抽象语法树差异
给定Eclipse中的以下代码:Java Eclipse抽象语法树差异,java,eclipse,diff,abstract-syntax-tree,compilationunit,Java,Eclipse,Diff,Abstract Syntax Tree,Compilationunit,给定Eclipse中的以下代码: import org.eclipse.jdt.core.dom.AST; import org.eclipse.jdt.core.dom.ASTParser; import org.eclipse.jdt.core.dom.CompilationUnit; public class Question { public static void main(String[] args) { String source = "class Bob
import org.eclipse.jdt.core.dom.AST;
import org.eclipse.jdt.core.dom.ASTParser;
import org.eclipse.jdt.core.dom.CompilationUnit;
public class Question {
public static void main(String[] args) {
String source = "class Bob {}";
ASTParser parser = ASTParser.newParser(AST.JLS3);
parser.setSource(source.toCharArray());
CompilationUnit result = (CompilationUnit) parser.createAST(null);
String source2 = "class Bob {public void MyMethod(){}}";
ASTParser parser2 = ASTParser.newParser(AST.JLS3);
parser2.setSource(source2.toCharArray());
CompilationUnit result2 = (CompilationUnit) parser2.createAST(null);
}
}
如何使用EclipseCompareAPI(org.Eclipse.Compare)查找AST差异?(这可以在插件之外完成吗?)
我正在看下面的API
任何人都可以指向示例代码(或API,但代码是首选)。鉴于Eclipse不进行AST差异,OP可能希望找到两个文件在忽略空格和注释的语言构造方面的差异。我们根据langauge结构(变量、表达式、语句、块、方法等)比较了两个源文件,并描述了这些元素在抽象编辑操作方面的差异(删除、复制、移动、重命名区域中的标识符等)免费执行此项工作:)
它还支持其他语言,如javascript。实际上,使用ASTNode的属性检查相等性很简单。在那之后,如何获得差异取决于你。检查代码示例以进行相等性测试:
public class ASTCompare {
@SuppressWarnings("unchecked")
static boolean equals(ASTNode left, ASTNode right) {
// if both are null, they are equal, but if only one, they aren't
if (left == null && right == null) {
return true;
} else if (left == null || right == null) {
return false;
}
// if node types are the same we can assume that they will have the same
// properties
if (left.getNodeType() != right.getNodeType()) {
return false;
}
List<StructuralPropertyDescriptor> props = left
.structuralPropertiesForType();
for (StructuralPropertyDescriptor property : props) {
Object leftVal = left.getStructuralProperty(property);
Object rightVal = right.getStructuralProperty(property);
if (property.isSimpleProperty()) {
// check for simple properties (primitive types, Strings, ...)
// with normal equality
if (!leftVal.equals(rightVal)) {
return false;
}
} else if (property.isChildProperty()) {
// recursively call this function on child nodes
if (!equals((ASTNode) leftVal, (ASTNode) rightVal)) {
return false;
}
} else if (property.isChildListProperty()) {
Iterator<ASTNode> leftValIt = ((Iterable<ASTNode>) leftVal)
.iterator();
Iterator<ASTNode> rightValIt = ((Iterable<ASTNode>) rightVal)
.iterator();
while (leftValIt.hasNext() && rightValIt.hasNext()) {
// recursively call this function on child nodes
if (!equals(leftValIt.next(), rightValIt.next())) {
return false;
}
}
// one of the value lists have additional elements
if (leftValIt.hasNext() || rightValIt.hasNext()) {
return false;
}
}
}
return true;
}
}
公共类{
@抑制警告(“未选中”)
静态布尔等于(ASTNode left,ASTNode right){
//如果两者都为空,则它们相等,但如果只有一个,则它们不相等
if(left==null&&right==null){
返回true;
}else if(左==null | |右==null){
返回false;
}
//如果节点类型相同,我们可以假设它们具有相同的
//性质
if(left.getNodeType()!=right.getNodeType()){
返回false;
}
列表道具=左
.StructurePropertiesForType();
for(StructuralPropertyDescriptor属性:props){
Object leftVal=left.getStructuralProperty(属性);
Object rightVal=right.getStructuralProperty(属性);
if(property.isSimpleProperty()){
//检查简单属性(基本类型、字符串等)
//正常相等
如果(!leftVal.等于(rightVal)){
返回false;
}
}else if(property.isChildProperty()){
//在子节点上递归调用此函数
如果(!等于((ASTNode)leftVal,(ASTNode)rightVal)){
返回false;
}
}else if(property.isChildListProperty()){
迭代器leftValIt=((Iterable)leftVal)
.iterator();
迭代器rightValIt=((Iterable)rightVal)
.iterator();
while(leftValIt.hasNext()&&righvalit.hasNext()){
//在子节点上递归调用此函数
如果(!等于(leftValIt.next(),rightValIt.next()){
返回false;
}
}
//其中一个值列表包含其他元素
if(leftValIt.hasNext()| | righvalit.hasNext()){
返回false;
}
}
}
返回true;
}
}
有开源解决方案吗?显然没有