Java 遍历AccessibilityNodeInfo时Android可访问性服务中的无限递归

我的应用程序的用户在几天后就遇到了崩溃。我能够在我的设备上收集logcat，并添加了一些调试输出。下面是正在发生的事情：

在我的

handleAccessibilityEvent（）

中，我调用

AccessibilityNodeInfo root = getRootInActiveWindow();
int eventWindowId = event.getWindowId();
if (ExistsNodeOrChildren(root, new WindowIdCondition(eventWindowId)))
{
}

它递归地遍历节点树：

private boolean ExistsNodeOrChildren(AccessibilityNodeInfo n, NodeCondition condition) 
{
    Log.d(_logTag, "ExistsNodeOrChildren" + n.toString());
    if (n == null) return false;
    if (condition.check(n))
        return true;
    for (int i = 0; i < n.getChildCount(); i++)
    {
        Log.d(_logTag, "ExistsNodeOrChildren child" + i);
        if (ExistsNodeOrChildren(n.getChild(i), condition))
            return true;
    }
    return false;
}

---

如果这不是由NodeCondition函数（您可能应该提供）中的其他递归引起的，并且这实际上是Android OS虚拟视图层次结构中的一个错误，那么您可以尝试通过执行以下操作来修复它：

private boolean ExistsNodeOrChildren(AccessibilityNodeInfo n, NodeCondition condition) {

    //For God sake do this first if you think n might actually be null!!!
    //Or just don't do it, and let n.toString() throw a NPE. (BAD IDEA)
    if (n == null) return false; 

    Log.d(_logTag, "ExistsNodeOrChildren" + n.toString());

    //NOTE: This could also cause recursion, you didn't provide this code.
    if (condition.check(n)) return true;

    for (int i = 0; i < n.getChildCount(); i++) {
        AccessibilityNodeInfo child = n.getChild(i);

        //Skip recursion the times n.getChild() returns n.
        //The check really is this simple, because we can only skip this
        //When the child is literally the same object, otherwise it might be
        //a node that has identical properties, which can happen.
        if (child != n) {
            Log.d(_logTag, "ExistsNodeOrChildren child" + i);
            if (ExistsNodeOrChildren(n.getChild(i), condition)) return true;
        } else {
            log.e("We should report this as a bug in the AOSP");
        }
    }

    return false;
}

private boolean ExistsNodeOrChildren（AccessibilityNodeInfo，NodeCondition条件）{
//看在上帝的份上，如果你认为n实际上可能是空的，请先这样做！！！
//或者干脆不做，让n.toString（）抛出一个NPE。（坏主意）
如果（n==null）返回false；
Log.d（_logTag，“ExistsNodeOrChildren”+n.toString（））；
//注意：这也可能导致递归，您没有提供此代码。
if（条件检查（n））返回true；
对于（int i=0；i

嘿，菲利普。从Chrome 1月25日的更新开始，bitwarden也出现了类似的问题。请参阅我发现的另一个可能相关的bug：次要注释（n.toString（））后跟n==null检查？真可疑！但最终是不相关的。

private boolean ExistsNodeOrChildren(AccessibilityNodeInfo n, NodeCondition condition) {

    //For God sake do this first if you think n might actually be null!!!
    //Or just don't do it, and let n.toString() throw a NPE. (BAD IDEA)
    if (n == null) return false; 

    Log.d(_logTag, "ExistsNodeOrChildren" + n.toString());

    //NOTE: This could also cause recursion, you didn't provide this code.
    if (condition.check(n)) return true;

    for (int i = 0; i < n.getChildCount(); i++) {
        AccessibilityNodeInfo child = n.getChild(i);

        //Skip recursion the times n.getChild() returns n.
        //The check really is this simple, because we can only skip this
        //When the child is literally the same object, otherwise it might be
        //a node that has identical properties, which can happen.
        if (child != n) {
            Log.d(_logTag, "ExistsNodeOrChildren child" + i);
            if (ExistsNodeOrChildren(n.getChild(i), condition)) return true;
        } else {
            log.e("We should report this as a bug in the AOSP");
        }
    }

    return false;
}