Puzzling behaviour of Py_IncRef

[Barry]

> On 25 Jan 2022, at 23:50, Tony Flury <tony.flury at btinternet.com> wrote:
> 
> 
> 
> 
>> On 25/01/2022 22:28, Barry wrote:
>> 
>>>> On 25 Jan 2022, at 14:50, Tony Flury via Python-list <python-list at python.org> wrote:
>>>> 
>>>> 
>>>>> On 20/01/2022 23:12, Chris Angelico wrote:
>>>>>>> On Fri, 21 Jan 2022 at 10:10, Greg Ewing <greg.ewing at canterbury.ac.nz> wrote:
>>>>>>> On 20/01/22 12:09 am, Chris Angelico wrote:
>>>>>>> At this point, the refcount has indeed been increased.
>>>>>>> 
>>>>>>>>           return self;
>>>>>>>>      }
>>>>>>> And then you say "my return value is this object".
>>>>>>> 
>>>>>>> So you're incrementing the refcount, then returning it without
>>>>>>> incrementing the refcount. Your code is actually equivalent to "return
>>>>>>> self".
>>>>>> Chris, you're not making any sense. This is C code, so there's no
>>>>>> way that "return x" can change the reference count of x.
>>>>> Yeah, I wasn't clear there. It was equivalent to *the Python code*
>>>>> "return self". My apologies.
>>>>> 
>>>>>  > The normal thing to do is to add a reference to whatever you're
>>>>>  > returning. For instance, Py_RETURN_NONE will incref None and then
>>>>>  > return it.
>>>>>  >
>>>>> 
>>>>> The OP understands that this is not a normal thing to do. He's
>>>>> trying to deliberately leak a reference for the purpose of diagnosing
>>>>> a problem.
>>>>> 
>>>>> It would be interesting to see what the actual refcount is after
>>>>> calling this function.
>>> After calling this without a double increment in the function the ref count is still only 1 - which means that the 'return self' effectively does a double decrement. My original message includes the Python code which calls this 'leaky' function and you can see that despite the 'leaky POC' doing an increment ref count drops back to one after the return.
>>> 
>>> You are right this is not a normal thing to do, I am trying to understand the behaviour so my library does the correct thing in all cases - for example - imagine you have two nodes in a tree :
>>> 
>>> A --- > B
>>> 
>>> And your Python code has a named reference to A, and B also maintains a reference to A as it's parent.
>>> 
>>> In this case I would expect A to have a reference count of 2 (counted as 3 through sys.getrefcount() - one for the named reference in the Python code - and one for the link from B back to A; I would also expect B to have a reference count here of 1 (just the reference from A - assuming nothing else referenced B).
>>> 
>>> My original code was incrementing the ref counts of A and B and then returning A. within the Python test code A had a refcount of 1 (and not the expected 2), but the refcount from B was correct as far as I could tell.
>>> 
>>> 
>>>> Yes, and that's why I was saying it would need a *second* incref.
>>>> 
>>>> ChrisA
>>> Thank you to all of you for trying to help - I accept that the only way to make the code work is to do a 2nd increment.
>>> 
>>> I don't understand why doing a 'return self' would result in a double decrement - that seems utterly bizzare behaviour - it obviously works, but why.
>> The return self in C will not change the ref count.
>> 
>> I would suggest setting a break point in your code and stepping out of the function and seeing that python’s code does to the ref count.
>> 
>> Barry
> Barry,
> 
> something odd is going on because the Python code isn't doing anything that would cause the reference count to go from 3 inside the C function to 1 once the method call is complete.
> 
> As far as I know the only things that impact the reference counts are : 
> 
> Increments due to assigning a new name or adding it to a container.
> Increment due to passing the object to a function (since that binds a new name) 
> Decrements due to deletion of a name
> Decrement due to going out of scope
> Decrement due to being removed from a container.
> None of those things are happening in the python code.
> 

Run python and your code under a debugger and check the ref count of the object as you step through the code.

Don’t just step through your code but also step through the C python code.
That will allow you to see how this works at a low level. 
Setting a watch point on the ref count will allow you run the code and just break as the ref count changes.

That is what I do when a see odd c api behaviour.

Barry
> As posted in the original message - immediately before the call to the C function/method sys.getrefcount reports the count to be 2 (meaning it is actually a 1).
> 
> Inside the C function the ref count is incremented and the Py_REFCNT macro reports the count as 3 inside the C function as expected (1 for the name in the Python code, 1 for the argument as passed to the C function, and 1 for the increment), so outside the function one would expect the ref count to now be 2 (since the reference caused by calling the function is then reversed).
> 
> However - Immediately outside the C function and back in the Python code sys.getrefcount reports the count to be 2 again - meaning it is now really 1. So that means that the refcount has been decremented twice in-between the return of the C function and the execution of the immediate next python statement. I understand one of those decrements - the parameter's ref count is incremented on the way in so the same object is decremented on the way out (so that calls don't leak references) but I don't understand where the second decrement is coming from.
> 
> Again there is nothing in the Python code that would cause that decrement - the decrement behavior is in the Python runtime.
> 
>>> 
>>> 
>>> -- 
>>> Anthony Flury
>>> email : anthony.flury at btinternet.com
>>> 
>>> -- 
>>> https://mail.python.org/mailman/listinfo/python-list
>>> 
> -- 
> Anthony Flury
> email : anthony.flury at btinternet.com