如何在Eclipse中获得超类节点？

Question

我这里有个密码：

public class TestOverride {
    int foo() {
        return -1;
    }
}

class B extends TestOverride {
    @Override
    int foo() {
        // error - quick fix to add "return super.foo();"   
    }
}

如你所见，我已经提到了这个错误。我正试图在eclipse中为此创建一个快速修复程序。但是我无法得到B类的超类节点，即类TestOverride。

我尝试了以下代码

if(selectedNode instanceof MethodDeclaration) {
    ASTNode type = selectedNode.getParent();
    if(type instanceof TypeDeclaration) {
        ASTNode parentClass = ((TypeDeclaration) type).getSuperclassType();
    }
}

在这里，我只把parentClass作为TestOverride。但是，当我检查它不是TypeDeclaration类型时，它也不是SimpleName类型。

我的查询是如何得到类TestOverride节点的？

编辑

  for (IMethodBinding parentMethodBinding :superClassBinding.getDeclaredMethods()){
     if (methodBinding.overrides(parentMethodBinding)){
        ReturnStatement rs = ast.newReturnStatement();
        SuperMethodInvocation smi = ast.newSuperMethodInvocation();
        rs.setExpression(smi);
        Block oldBody = methodDecl.getBody();
        ListRewrite listRewrite = rewriter.getListRewrite(oldBody, Block.STATEMENTS_PROPERTY);
        listRewrite.insertFirst(rs, null);
}

Answer 1

您必须使用bindings。要使绑定可用，这意味着resolveBinding()不返回null，我发布的possibly additional steps是必要的。

要处理绑定，这个访问者应该有助于正确的方向：

class TypeHierarchyVisitor extends ASTVisitor {
    public boolean visit(MethodDeclaration node) {
        // e.g. foo()
        IMethodBinding methodBinding = node.resolveBinding();

        // e.g. class B
        ITypeBinding classBinding = methodBinding.getDeclaringClass();

        // e.g. class TestOverride
        ITypeBinding superclassBinding = classBinding.getSuperclass();
        if (superclassBinding != null) {
            for (IMethodBinding parentBinding: superclassBinding.getDeclaredMethods()) {
                if (methodBinding.overrides(parentBinding)) {
                    // now you know `node` overrides a method and
                    // you can add the `super` statement
                }
            }
        }
        return super.visit(node);
    }
}