如何查找ASTNode的所有子节点（儿童和儿童孩子）

Question

我试图通过获取ExpressionStatements并返回它们的子节点及其子节点来获取AST节点的所有子节点，但算法陷入了第一次ExpStat和我找不到原因。

首先我创建了一个访客功能，找到我的类的所有ExpressionStatements，然后调用该函数找到你的孩子

private void analyseClass(ICompilationUnit classe) throws JavaModelException { 
    // ICompilationUnit unit == class 
    // now create the AST for the ICompilationUnits 
    CompilationUnit parse = parse(classe); 

    // Calls the method for visit node in AST e return your information 
    ExpressionStatementVisitor visitor = new ExpressionStatementVisitor(); 
    parse.accept(visitor); 

    // Write in the screen: ExpressionStatement and your type next 
    for (ExpressionStatement method : visitor.getExpression()) { 
     //String t = null; 

     // 32 -> METHOD_INVOCATION type 
     if (method.getExpression().getNodeType() == 32) { 
      getChildren(method); 
      results.append("\n\n"); 
     } 

     // 48 -> SUPER_METHOD_INVOCATION type 
     else if (method.getExpression().getNodeType() == 48) { 
      // results.append("\n SuperMethodInvocation: " + t); 
      //getChildren(method); 
      //results.append("\n\n"); 
     } else { 
      //getChildren(method); 
      //results.append("\n\n"); 
     } 
    } 
} 

找到孩子递归函数：

public static void getChildren(ASTNode node) { 
    if (node != null) { 
     List<ASTNode> children = new ArrayList<ASTNode>(); 
     List list = node.structuralPropertiesForType(); 
     for (int i = 0; i < list.size(); i++) { 
      Object child = node.getStructuralProperty((StructuralPropertyDescriptor) list.get(i)); 
      if (child instanceof ASTNode) { 
       children.add((ASTNode) child); 
      }    
      if (children.get(0) != null) { 
       String c = children.toString(); 
       results.append("Children Node: " + c + "\n"); 
       getChildren(children.get(0)); 
      } 
     } 
    } else { 
     return; 
    }  
} 

我们假设在课堂上有：

a.getTheDataA().getTheDataB().getTheDataC().getTheData(); 
b.getTheDataA().getTheDataB().getTheDataC().getTheData(); 
c.getTheE(a,b).getTheF(getTheDataB).getTheH(); 

getChildren function rea ds只有a.getTheDataA（）。getTheDataB（）。getTheDataC（）。getTheData（）;并返回他的孩子和儿童的孩子是这样的：

print screen

我卡在这一天，我需要从我所看到的递归

Answer 1

帮助，你永远只能得到的第一个元素children，我认为你需要把这个语句检查一下，看看children元素是否为空，并将其分别放入for循环中，并检查其中的每个元素。

喜欢的东西：

public static void getChildren(ASTNode node) { 
    if (node != null) { 
     List<ASTNode> children = new ArrayList<ASTNode>(); 
     List list = node.structuralPropertiesForType(); 
     for (int i = 0; i < list.size(); i++) { 
      Object child = node.getStructuralProperty((StructuralPropertyDescriptor) list.get(i)); 
      if (child instanceof ASTNode) { 
       children.add((ASTNode) child); 
      }    
     } 
     for(ASTNode node : children){ 
      if (node != null) { 
       String c = children.toString(); 
       results.append("Children Node: " + c + "\n"); 
       getChildren(node); 
      } 
     } 
    }else { 
     return; 
    }  
} 

我还没有运行的代码，但我认为问题是，你只能得到解决的children

Answer 2

第一要素！

public static int getChildren(ASTNode node,int n) { 
    int cont = n; 
    String compara = "[]"; 

    List<ASTNode> children = new ArrayList<ASTNode>(); 
    @SuppressWarnings("rawtypes") 
    List list = node.structuralPropertiesForType(); 

    for (int i = 0; i < list.size(); i++) { 
     Object child = node.getStructuralProperty((StructuralPropertyDescriptor)list.get(i)); 
     if (child instanceof ASTNode) { 
      children.add((ASTNode) child); 
     } 
    } 

    String teste = children.toString(); 

    // Se a string do filho for igual a [] -> CHEGOU AO FIM 
    //e retorna resultado do contador para analyseClass 
    if (teste.equals(compara)) { 
     results.append("NMCS = "+cont+"\n"); 
     return cont; 
    } 

    // Aumenta o contador se o nó filho for MethodInvocation ou 
    //SuperMethodInvocation 
    if (node.getNodeType() == 32) { 
     cont++; 
    } else if (node.getNodeType() == 48) { 
     cont++; 
    } 

    // Recursão para encontrar próximo nó (filho do filho) 
    return getChildren(children.get(0),cont);}