So, is the granularity that of seq realization -- individual [first & rest]
cells for (iterate inc 0), single chunks for (range), etc.? I'd appreciate
a straight, direct, yes-or-no answer to that question.


On Mon, Jun 24, 2013 at 11:03 PM, Timothy Baldridge <tbaldri...@gmail.com>wrote:

> It corresponds to the execution of the LazySeq fn. That fn will be called
> once and only once, the rest of the data in the object is immutable and
> side-effect free and therefore does not need to be synchronized.
>
> Timothy
>
>
> On Mon, Jun 24, 2013 at 8:51 PM, Cedric Greevey <cgree...@gmail.com>wrote:
>
>> I'm familiar with what "synchronized Type foo (args)" does -- my last
>> question was more about what aspect of a lazy seq the object with the
>> method corresponds to. Cons cell or similar subunit? I could read half of
>> clojure.lang, learn how all the various types of seq (Cons, LazySeq,
>> ChunkedSeq, etc...) work under the hood, and thereby eventually figure it
>> out, but it's probably a lot fewer man-hours of work for me to ask someone
>> who's already intimately familiar with that codebase and for him to
>> answer...
>>
>>
>>
>> On Mon, Jun 24, 2013 at 10:32 PM, Timothy Baldridge <tbaldri...@gmail.com
>> > wrote:
>>
>>> Reading the LazySeq.java file should make this all clear, but yes, no
>>> race conditions.
>>>
>>>
>>> https://github.com/clojure/clojure/blob/master/src/jvm/clojure/lang/LazySeq.java#L37
>>>
>>> Synchronized methods basically lock the current instance of the object
>>> while the method runs, so it is impossible for two threads to execute the
>>> lazy seq fn at the same time.
>>>
>>> Timothy
>>>
>>>
>>> On Mon, Jun 24, 2013 at 4:17 PM, Cedric Greevey <cgree...@gmail.com>wrote:
>>>
>>>> Ah, thanks. The locking granularity is local to the cons cell (or
>>>> analogue; first/rest pair) being realized, I hope? So one thread actually
>>>> calculates an element and neither call returns until it's calculated, and
>>>> then both promptly return the calculated value, but threads realizing other
>>>> lazy seqs or crawling along earlier parts of the same one don't get
>>>> blocked? (And given they can share tails, how would "same one" even be
>>>> defined anyway?)
>>>>
>>>>
>>>> On Mon, Jun 24, 2013 at 5:56 PM, Nicola Mometto <brobro...@gmail.com>wrote:
>>>>
>>>>>
>>>>> Realizing a lazy-seq is done through a synchronized method see:
>>>>>
>>>>> https://github.com/clojure/clojure/blob/master/src/jvm/clojure/lang/LazySeq.java#L37
>>>>>
>>>>> No race conditions.
>>>>>
>>>>> Cedric Greevey writes:
>>>>>
>>>>> > What, precisely, happens if two threads sharing a reference to a
>>>>> single
>>>>> > lazy sequence try to realize the same element at the same time? If
>>>>> the
>>>>> > sequence is completely pure and deterministic, so any attempt to
>>>>> realize a
>>>>> > particular element will produce a single particular value
>>>>> consistently
>>>>> > (unlike, say, (repeatedly rand) or a file-seq where relevant parts
>>>>> of the
>>>>> > filesystem are being concurrently modified), is the worst-case
>>>>> scenario
>>>>> > that the two threads will redundantly perform the same calculation,
>>>>> with no
>>>>> > effect other than a minor hit to performance and, in particular, no
>>>>> effect
>>>>> > on the program semantics?
>>>>> >
>>>>> > --
>>>>>
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>>>
>>>
>>>
>>> --
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>>> zero–they had no way to indicate successful termination of their C
>>> programs.”
>>> (Robert Firth)
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>
>
>
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> “One of the main causes of the fall of the Roman Empire was that–lacking
> zero–they had no way to indicate successful termination of their C
> programs.”
> (Robert Firth)
>
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