Hi Eric,

On 10/2/2017 9:40 PM, Eric Dobson wrote:
George: I don't see that invariant for java in anything that I search for on Java finalizers. Do you have a reference? In particular java only has one finalizer per object (the finalize method) and across objects the references I found seem to imply that there is no guaranteed order?

https://en.wikipedia.org/wiki/Finalization#Connection_with_finally seems to imply no ordering across finalizers.

What I recall was that members were guaranteed to be finalized (if needed) before their parent object - the reverse of the initialization process.  This did not necessary conflict with "objects may be finalized in any order" ... it simply meant that when the parent was to be finalized, some or all of the members also needing finalization might have already been done.

Unfortunately I don't recall exactly where I saw it ... I'm pretty sure it was in a paper on JVM internals and so was maybe about what was actually done vs what was guaranteed by specification.  And it was quite a while ago that I was studying virtual machines, so it may no longer be true of current implementations.  I'll have to look around and see if I can find it again.

Even so, my question to Matthew was - would something like this be workable for Racket?  I don't know how much additional bookkeeping would be required.

Matthew: Ok, I'll just file a bug to track the issue. I'm trying to do the best available solution following standard practices, and not actually that worried about the memory leak for my program.

On Mon, Oct 2, 2017 at 1:48 PM, George Neuner <gneun...@comcast.net <mailto:gneun...@comcast.net>> wrote:

    IIRC, the JVM executes finalizers in reverse order of registration
    - expecting that, in your example, A would have been allocated
    before B.  Is that something that would work here?  [He asks naively.]


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