如何找到一个ASTNode的所有子节点（children和childrens children）

Question

我试图通过获取 ExpressionStatements 并返回它们的 child 及其子 children 来获取 AST 节点的所有子节点，但是算法卡在第一个 ExpStat 和我找不到原因。

首先我创建了一个访问者函数来查找我的 class 的所有 ExpressionStatements，然后我调用该函数来查找您的 children

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");
        }
    }
}

递归查找children的函数：

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; 
    }       
}

假设 class 有：

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

getChildren 函数只读取 a.getTheDataA().getTheDataB().getTheDataC().getTheData(); returns 他的 children 和 children children 像这样：

print screen

有一天我卡在这上面了，我需要递归方面的帮助

Answer 1

据我所知，你只能得到 children 的第一个元素，我认为你需要将检查 children 元素是否不为 null 的语句取出到一个单独的 for 循环，并检查其中的每个元素。

类似于：

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; 
    }       
}

我没有运行代码，但我认为问题在于您只能获得 children

的第一个元素

Answer 2

已解决！

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);}

如何找到一个ASTNode的所有子节点（children和childrens children）

How to find all sub nodes (children and childrens children) of an ASTNode

java

eclipse

recursion

eclipse-rcp

abstract-syntax-tree