Sure, slot does have a value, I just didn't include it in the code. 
Something like:

Local<Value> slot = String::NewFromUtf8(isolate, "MyBeautifulSlot", 
NewStringType::kNormal).ToLocalChecked();

Cheers!

On Thursday, June 21, 2018 at 2:27:00 PM UTC+2, Zac Hansen wrote:
>
> You're dereferencing a "super pointer" to get to a "pointer", hence * not 
> &.   You can't "go back" because the local/global<T> represents an 
> "reference count" to the object which must be known to the JS runtime.   
>
> As for p0 and p1, have you tried setting slot to a fixed string value 
> before using it as a key for storing/lookup?   I don't know what the 
> expected behavior of using an empty value as a key into an object is.
>
> These are all just guesses - if someone else answers differently, I'm 
> probably wrong.
>
>
> On Thursday, June 21, 2018 at 5:07:50 AM UTC-7, Gonzalo Diethelm wrote:
>>
>> Note to self: this might be related to Local<Object> vs Global<Object> 
>> (or Persistent<Object>? so many names...)
>>
>> Need to look into that.
>>
>> On Thursday, June 21, 2018 at 7:58:37 AM UTC+2, Gonzalo Diethelm wrote:
>>>
>>> I run the following JS code in the Chrome console:
>>>
>>> // Version 67.0.3396.87 (Official Build) (64-bit)
>>>
>>> var x = [1, 2, {"foo": 11}];
>>> x[2].bar = x;
>>>
>>> Now from C++ code, I get ahold of x as a Local<Object>, and wish to 
>>> traverse the whole structure; for the sake of the example, let's say I am 
>>> converting it into serialized data (I know I can use JSON.stringify() to do 
>>> this, serializing is just an example to clarify ideas).  My question is, 
>>> how can I keep track of the nodes in the structure  that I have already 
>>> seen, and their associated serialized value, so that I can avoid an 
>>> infinite traversal?
>>>
>>> It seems to me doing this would require a way to get a unique identity 
>>> for each node, so that the C++ code can do something similar to this:
>>>
>>> typedef map<NodeId, NodeData> NodeMap;
>>> NodeMap seen;
>>> ...
>>> Local<Object> node = current.GetNextChild();
>>> NodeId id = node.GetUniqueId();
>>> NodeMap::iterator k = seen.find(id);
>>> NodeData data;
>>> if (k != seen.end()) {
>>>     // node already seen, reuse its serialization
>>>     data = k->first;
>>> } else {
>>>     // first time we see node, serialize and remember
>>>     data = node.Serialize(); // recurses
>>>     seen[id] = data;
>>> }
>>>
>>> The specific question is: what type could be NodeId, and how do I get 
>>> the equivalent of GetUniqueId()?
>>>
>>> I am very tempted to ask for a way to get a raw void* to each node, but 
>>> I guess any way of doing this is fine, as long as I can get a unique id 
>>> that is stable while I'm traversing the data.  For these reasons, 
>>> GetIdentityHash() does not seem to fit the bill: "*The return value 
>>> will never be 0. Also, it is not guaranteed to be unique.*"
>>>
>>> Incidentally, If I try to use JSON.stringify for my data, I get this:
>>>
>>> JSON.stringify(x)
>>> VM170:1 Uncaught TypeError: Converting circular structure to JSON
>>>     at JSON.stringify (<anonymous>)
>>>     at <anonymous>:1:6
>>>
>>> This is taken care of here in the V8 code:
>>>
>>> JsonStringifier::Result JsonStringifier::StackPush(Handle<Object> object
>>> ) {
>>> ...
>>>     // member stack_ is: Handle<JSArray> stack_;
>>>     int length = Smi::ToInt(stack_->length());
>>>     FixedArray* elements = FixedArray::cast(stack_->elements());
>>>     for (int i = 0; i < length; i++) {
>>>         FixedArray* elements = FixedArray::cast(stack_->elements());
>>>         if (elements->get(i) == *object) {
>>>             // boom
>>>         }
>>>     }
>>> }
>>>
>>> So, operator*() in a Handle<Object> gives me a unique id? Which is the 
>>> type for this? Can I store that in a C++ map? Is it stable (enough)?
>>>
>>> Thanks!
>>>
>>

-- 
-- 
v8-users mailing list
v8-users@googlegroups.com
http://groups.google.com/group/v8-users
--- 
You received this message because you are subscribed to the Google Groups 
"v8-users" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to v8-users+unsubscr...@googlegroups.com.
For more options, visit https://groups.google.com/d/optout.

Reply via email to