问题
given:
-record(foo, {a, b, c}).
I do something like this:
Thing = #foo{a={1,2}, b={3,4}, c={5,6}},
Thing1 = Thing#foo{a={7,8}}.
From a semantic view, Thing and Thing1 are unique entities. However, from a language implementation standpoint, making a full copy of Thing to generate Thing1 would be intensely wasteful. For example, if the record were a megabyte in size and I made a thousand "copies," each modifying a couple of bytes, I've just burned a gigabyte. If the internal structure kept track of a representation of the parent structure and each derivative marked up that parent in a way that indicated its own change but preserved everyone elses' versions, the derivates could be created with a minimum of memory overhead.
My question is this: is erlang doing anything clever - internally - to keep the overhead of the usual erlang scribble;
Thing = #ridiculously_large_record,
Thing1 = make_modified_copy(Thing),
Thing2 = make_modified_copy(Thing1),
Thing3 = make_modified_copy(Thing2),
Thing4 = make_modified_copy(Thing3),
Thing5 = make_modified_copy(Thing4)
...to a minimum?
I ask because there would be a number of changes to the way that I did cross-process communications if this were the case.
回答1:
The exact workings of the garbage collection and memory allocation is only known to a few. Thankfully, they are very happy to share their knowledge and the following is based on what I have learnt from the erlang-questions mailing list and by discussing with OTP developers.
When messaging between processes, the content is always copied as there is no shared heap between processes. The only exception is binaries bigger than 64 bytes, where only a reference is copied.
When executing code in one process, only parts are updated. Let's analyze tuples, as that is the example you provided.
A tuple is actually a structure that keeps references to the actual data somewhere on the heap (except for small integers and maybe one more data type which I can't remember). When you update a tuple, using for example setelement/3, a new tuple is created with the given element replaced, however for all other elements only the reference is copied. There is one exception which I have never been able to take advantage of.
The garbage collector keeps track of each tuple and understands when it is safe to reclaim any tuple that is no longer used. It might be that the data referenced by the tuple is still in use, in which case the data itself is not collected.
As always, Erlang gives you some tools to understand exactly what is going on. The efficiency guide details how to use erts_debug:size/1 and erts_debug:flat_size/1 to understand the size of the data structure when used internally in a process and when copied. The trace tools also allows you to understand when, what and how much was garbage collected.
回答2:
The record foo is of arity four (holding four words), but the whole structure is 14 words in size. Any immediate (pids, ports, small integers, atoms, catch and nil) can be stored directly in the tuples array. Any other term which can't fit into a word, such as other tuples, are not stored directly but referenced by boxed pointers (a boxed pointer is an erlang term with a forwarding address to the real eterm ... just internals).
In your case a new tuple of same arity is created and the atom foo and all the pointers are copied from the previous tuple except for index two, a, which points to the new tuple {7,8} which constitutes 3 words. In all 5 + 3 new words are created on the heap and only 3 words are copied from the old tuple the other 9 words are not touched.
Excessively large tuples are not recommended. When updating a tuple, the whole tuple, i.e the array and not the deep content, needs to copied and then updated in other to preserve a persistent data structure. This will also generate increased garbage, forcing the garbage collector to heat up which also hurts performance. The dict and array modules avoids using large tuples for this reason and have a shallow tuple tree instead.
回答3:
I can definitely verify what people have already pointed out:
- a record is just a tuple with the record name as the first element and all the fields just the following tuple element
- when an element of a tuple is changed, updating a field in a record in your case, only the top level tuple is new, all the elements are just reused
This works just because we have immutable data. So in your example each time you update a value in a #foo record none of the data in the elements are copied and only a new 4-element tuple (5 words) is created. Erlang will never does a deep copy in this type of operation or when passing arguments in function calls.
回答4:
In conclusion:
Thing = #foo{a={1,2}, b={3,4}, c={5,6}},
Thing1 = Thing#foo{a={7,8}}.
Here, if Thing is not used again, it will probably be updated in place and copying of the tuple will be avoided, as the Efficiency Guide says. (tuple and record syntax is complied into something like setelement, I think)
Thing = #ridiculously_large_record,
Thing1 = make_modified_copy(Thing),
Thing2 = make_modified_copy(Thing1),
...
Here the tuples are actually copied every time.
I guess that it would be theoretically possible make an interesting optimization to this. If the compiler could perform escape analysis on the return value of make_modified_copy and detect that the only reference to it is the one returned, in could save this information about the function. When it encounter a call the that function it would know that it is safe to modify the return value in place.
This would only be possible to do on inter module calls, because of the code replace feature.
Maybe one day we will have it.
来源：https://stackoverflow.com/questions/7207895/does-erlang-implement-record-copy-and-modify-in-any-clever-way