COPY and SWAP idiom

https://stackoverflow.com/questions/3279543/what-is-the-copy-and-swap-idiom

swap and move

Why do we need the copy-and-swap idiom?

Any class that manages a resource (a wrapper, like a smart pointer) needs to implement The Big Three. While the goals and implementation of the copy-constructor and destructor are straightforward, the copy-assignment operator is arguably the most nuanced and difficult. How should it be done? What pitfalls need to be avoided?

The copy-and-swap idiom is the solution, and elegantly assists the assignment operator in achieving two things: avoiding code duplication, and providing a strong exception guarantee.

How does it work?

Conceptually, it works by using the copy-constructor's functionality to create a local copy of the data, then takes the copied data with a swap function, swapping the old data with the new data. The temporary copy then destructs, taking the old data with it. We are left with a copy of the new data.

In order to use the copy-and-swap idiom, we need three things: a working copy-constructor, a working destructor (both are the basis of any wrapper, so should be complete anyway), and a swapfunction.

A swap function is a non-throwing function that swaps two objects of a class, member for member. We might be tempted to use std::swap instead of providing our own, but this would be impossible; std::swap uses the copy-constructor and copy-assignment operator within its implementation, and we'd ultimately be trying to define the assignment operator in terms of itself!

(Not only that, but unqualified calls to swap will use our custom swap operator, skipping over the unnecessary construction and destruction of our class that std::swap would entail.)

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An in-depth explanation

The goal

Let's consider a concrete case. We want to manage, in an otherwise useless class, a dynamic array. We start with a working constructor, copy-constructor, and destructor:

This class almost manages the array successfully, but it needs operator= to work correctly.

A failed solution

Here's how a naive implementation might look:

And we say we're finished; this now manages an array, without leaks. However, it suffers from three problems, marked sequentially in the code as (n).

1. The first is the self-assignment test. This check serves two purposes: it's an easy way to prevent us from running needless code on self-assignment, and it protects us from subtle bugs (such as deleting the array only to try and copy it). But in all other cases it merely serves to slow the program down, and act as noise in the code; self-assignment rarely occurs, so most of the time this check is a waste. It would be better if the operator could work properly without it.
2. The second is that it only provides a basic exception guarantee. If new int[mSize] fails, *this will have been modified. (Namely, the size is wrong and the data is gone!) For a strong exception guarantee, it would need to be something akin to:

3. The code has expanded! Which leads us to the third problem: code duplication. Our assignment operator effectively duplicates all the code we've already written elsewhere, and that's a terrible thing.

In our case, the core of it is only two lines (the allocation and the copy), but with more complex resources this code bloat can be quite a hassle. We should strive to never repeat ourselves.

(One might wonder: if this much code is needed to manage one resource correctly, what if my class manages more than one? While this may seem to be a valid concern, and indeed it requires non-trivial try/catch clauses, this is a non-issue. That's because a class should manage one resource only!)

A successful solution

As mentioned, the copy-and-swap idiom will fix all these issues. But right now, we have all the requirements except one: a swap function. While The Rule of Three successfully entails the existence of our copy-constructor, assignment operator, and destructor, it should really be called "The Big Three and A Half": any time your class manages a resource it also makes sense to provide a swap function.

We need to add swap functionality to our class, and we do that as follows†:

(Here is the explanation why public friend swap.) Now not only can we swap our dumb_array's, but swaps in general can be more efficient; it merely swaps pointers and sizes, rather than allocating and copying entire arrays. Aside from this bonus in functionality and efficiency, we are now ready to implement the copy-and-swap idiom.

Without further ado, our assignment operator is:

And that's it! With one fell swoop, all three problems are elegantly tackled at once.

Why does it work?

We first notice an important choice: the parameter argument is taken by-value. While one could just as easily do the following (and indeed, many naive implementations of the idiom do):

We lose an important optimization opportunity. Not only that, but this choice is critical in C++11, which is discussed later. (On a general note, a remarkably useful guideline is as follows: if you're going to make a copy of something in a function, let the compiler do it in the parameter list.‡)

Either way, this method of obtaining our resource is the key to eliminating code duplication: we get to use the code from the copy-constructor to make the copy, and never need to repeat any bit of it. Now that the copy is made, we are ready to swap.

Observe that upon entering the function that all the new data is already allocated, copied, and ready to be used. This is what gives us a strong exception guarantee for free: we won't even enter the function if construction of the copy fails, and it's therefore not possible to alter the state of *this. (What we did manually before for a strong exception guarantee, the compiler is doing for us now; how kind.)

At this point we are home-free, because swap is non-throwing. We swap our current data with the copied data, safely altering our state, and the old data gets put into the temporary. The old data is then released when the function returns. (Where upon the parameter's scope ends and its destructor is called.)

Because the idiom repeats no code, we cannot introduce bugs within the operator. Note that this means we are rid of the need for a self-assignment check, allowing a single uniform implementation of operator=. (Additionally, we no longer have a performance penalty on non-self-assignments.)

And that is the copy-and-swap idiom.

What about C++11?

The next version of C++, C++11, makes one very important change to how we manage resources: the Rule of Three is now The Rule of Four (and a half). Why? Because not only do we need to be able to copy-construct our resource, we need to move-construct it as well.

Luckily for us, this is easy:

What's going on here? Recall the goal of move-construction: to take the resources from another instance of the class, leaving it in a state guaranteed to be assignable and destructible.

So what we've done is simple: initialize via the default constructor (a C++11 feature), then swap with other; we know a default constructed instance of our class can safely be assigned and destructed, so we know other will be able to do the same, after swapping.

(Note that some compilers do not support constructor delegation; in this case, we have to manually default construct the class. This is an unfortunate but luckily trivial task.)

Why does that work?

That is the only change we need to make to our class, so why does it work? Remember the ever-important decision we made to make the parameter a value and not a reference:

Now, if other is being initialized with an rvalue, it will be move-constructed. Perfect. In the same way C++03 let us re-use our copy-constructor functionality by taking the argument by-value, C++11 will automatically pick the move-constructor when appropriate as well. (And, of course, as mentioned in previously linked article, the copying/moving of the value may simply be elided altogether.)

And so concludes the copy-and-swap idiom.

Lock free programming:

https://preshing.com/20120612/an-introduction-to-lock-free-programming/

memory ordering:
https://bartoszmilewski.com/2008/12/01/c-atomics-and-memory-ordering/

Kernel tuning:

cat /proc/cmdline
BOOT_IMAGE=/vmlinuz-3.10.0-123.el7.x86_64 root=UUID=d34860b6-1642-4a58-9a45-80e398065f28 ro vconsole.keymap=us crashkernel=auto vconsole.font=latarcyrheb-sun16 rhgb quiet console=ttyS1 intel_idle.max_cstate=0 processor.max_cstate=0 idle=poll hugepages=16384 isolcpus=1-8,10-35 nohz_full=1-8,10-35 rcu_nocbs=1-8,10-35 thermal.off=1 gbpages nosoftlockup

software tuning

crack inter

dvt-nj01-sv104:messages $ wireshark -X lua_script:/home/enguang/work/git/tng/fdk/yaml/messages/vfm_dissector.lua&

https://github.com/Open-Markets-Initiative/wireshark-lua

Illustrated Guide to Monitoring and Tuning the Linux Networking Stack: Receiving Data:

Tuning receive buf

memory allocation

First, some clarification. You asked: ... my program a.out terminates already, there is no other process that is maintaining this memory cache - or, is there one?

Everything we are talking about is within the lifetime of a single process: the process always returns all allocated memory when it exits. There is no cache that outlives the process¹. The memory is returned even without any help from the runtime allocator: the OS simply "takes it back" when the process is terminated. So there is no system-wide leak possible from terminated applications with normal allocations.

Now what Valgrind is reporting is memory that is in use at the moment the process terminated, but before the OS cleans everything up. It works at the runtime library level, and not at the OS level. So it's saying "Hey, when the program finished, there were 72,000 bytes that hadn't been returned to the runtime" but an unstated implication is that "these allocations will be cleaned up shortly by the OS".

The Underlying Questions

The code and Valgrind output shown doesn't really correlate well with the titular question, so let's break them apart. First we'll just try to answer the questions you asked about allocators: why they exist and why they don't generally don't immediately return freed memory to the OS, ignoring the example.

You asked:

1) Why keep them in an internal cache? If it is for speed, how is it faster? Yes, the OS needs to maintain a data structure to keep track of memory allocation, but this the maintainer of this cache also needs to do so.

This is sort of two questions in one: one is why bother having a userland runtime allocator at all, and then the other one is (perhaps?) why don't these allocators immediately return memory to the OS when it is freed. They are related, but let's tackle them one at a time.

Why Runtime Allocators Exist

Why not just rely on the OS memory allocation routines?

• Many operating systems, including most Linux and other Unix-like operating systems, simply don't have an OS system call to allocate and free arbitrary blocks of memory. Unix-alikes offer brk which only grows or shrinks one contiguous block of memory - you have no way to "free" arbitrary earlier allocations. They also offer mmap which allows you to independently allocate and free chunks of memory, but these allocate on a PAGE_SIZE granularity, which on Linux is 4096 bytes. So if you want request 32 bytes, you'll have to waste 4096 - 32 == 4064 bytes if you don't have your own allocator. On these operating systems you practically need a separate memory allocation runtime which turns these coarse-grained tools into something capable of efficiently allocating small blocks.
  Windows is a bit different. It has the HeapAlloc call, which is part of the "OS" and does offer malloc-like capabilities of allocating and freeing arbitrarily sized chunks of memory. With somecompilers then, malloc is just implemented as a thin wrapper around HeapAlloc (the performance of this call has improved greatly in recent Windows versions, making this feasible). Still, while HeapAlloc is part of the OS it isn't implemented in the kernel - it is also mostly implemented in a user-mode library, managing a list of free and used blocks, with occasional kernel calls to get chunks of memory from the kernel. So it is mostly malloc in another disguise and any memory it is holding on to is also not available to any other processes.
• Performance! Even if there were appropriate kernel-level calls to allocate arbitrary blocks of memory, the simple overhead roundtrip to the kernel is usually hundreds of nanoseconds or more. A well-tuned malloc allocation or free, on other hand, is often only a dozen instructions and may complete in 10 ns or less. On top of that, system calls can't "trust their input" and so must carefully validate parameters passed from user-space. In the case of free this means that it much check that the user passed a pointer which is valid! Most runtime free implements simply crash or silently corrupt memory since there is no responsibility to protect a process from itself.
• Closer link to the rest of the language runtime. The functions you use to allocate memory in C++, namely new, malloc and friends, are part of an defined by the language. It is then entirely natural to implement them as part of the runtime that implements the rest of the language, rather than the OS which is for the most part language-agnostic. For example, the language may have specific alignment requirements for various objects, which can best be handled by language aware allocators. Changes to the language or compiler might also imply necessary changes to the allocation routines, and it would be a tough call to hope for the kernel to be updated to accommodate your language features!

Why Not Return Memory to the OS

Your example doesn't show it, but you asked and if you wrote a different test you would probably find that after allocating and then freeing a bunch of memory, your processes resident set size and/or virtual size as reported by the OS might not decrease after the free. That is, it seems like the process holds on to the memory even though you have freed it. This is in fact true of many mallocimplementations. First, note that this is not a leak per se - the unreturned memory is still available to the process that allocated it, even if not to other processes.

Why do they do that? Here are some reasons:

1. The kernel API makes it hard. For the old-school brk and sbrk system calls, it simply isn't feasible to return freed memory unless it happens to be at the end of very last block allocated from brk or sbrk. That's because the abstraction offered by these calls is a single large contiguous region that you can only extend from one end. You can't hand back memory from the middle of it. Rather than trying to support the unusual case where all the freed memory happens to be at the end of brk region, most allocators don't even bother.
   The mmap call is more flexible (and this discussion generally applies also to Windows where VirtualAlloc is the mmap equivalent), allowing you to at least return memory at a page granularity - but even that is hard! You can't return a page until all allocations that are part of that page are freed. Depending on the size and allocation/free pattern of the application that may be common or uncommon. A case where it works well is for large allocations - greater than a page. Here you're guaranteed to be able to free most of the allocation if it was done via mmap and indeed some modern allocators satisfy large allocations directly from mmap and free them back to the OS with munmap. For glibc (and by extension the C++ allocation operators), you can even control this threshold:

   M_MMAP_THRESHOLD
     For allocations greater than or equal to the limit specified
     (in bytes) by M_MMAP_THRESHOLD that can't be satisfied from
     the free list, the memory-allocation functions employ mmap(2)
     instead of increasing the program break using sbrk(2).
   
     Allocating memory using mmap(2) has the significant advantage
     that the allocated memory blocks can always be independently
     released back to the system.  (By contrast, the heap can be
     trimmed only if memory is freed at the top end.)  On the other
     hand, there are some disadvantages to the use of mmap(2):
     deallocated space is not placed on the free list for reuse by
     later allocations; memory may be wasted because mmap(2)
     allocations must be page-aligned; and the kernel must perform
     the expensive task of zeroing out memory allocated via
     mmap(2).  Balancing these factors leads to a default setting
     of 128*1024 for the M_MMAP_THRESHOLD parameter.

   So by default allocations of 128K or more will be allocated by the runtime directly from the OS and freed back to the OS on free. So sometimes you will see the behavior you might have expected is always the case.
2. Performance! Every kernel call is expensive, as described in the other list above. Memory that is freed by a process will be needed shortly later to satisfy another allocation. Rather than trying to return it to the OS, a relatively heavyweight operation, why not just keep it around on a free listto satisfy future allocations? As pointed out in the man page entry, this also avoids the overhead of zeroing out all the memory returned by the kernel. It also gives the best chance of good cache behavior since the process is continually re-using the same region of the address space. Finally, it avoids TLB flushes which would be imposed by munmap (and possibly by shrinking via brk).
3. The "problem" of not returning memory is the worst for long-lived processes that allocate a bunch of memory at some point, free it and then never allocate that much again. I.e., processes whose allocation high-water mark is larger than their long term typical allocation amount. Most processes just don't follow that pattern, however. Processes often free a lot of memory, but allocate at a rate such that their overall memory use is constant or perhaps increasing. Applications that do have the "big then small" live size pattern could perhaps force the issue with malloc_trim.
4. Virtual memory helps mitigate the issue. So far I've been throwing around terms like "allocated memory" without really defining what it means. If a program allocates and then frees 2 GB of memory and then sits around doing nothing, is it wasting 2 GB of actual DRAM plugged into your motherboard somewhere? Probably not. It is using 2 GB of virtual address space in your process, sure, but virtual address space is per-process, so that doesn't directly take anything away from other processes. If the process actually wrote to the memory at some point, it would be allocated physical memory (yes, DRAM) - after freeing it, you are - by definition - no longer using it. At this point the OS may reclaim those physical pages by use for someone else.
   Now this still requires you have swap to absorb the dirty not-used pages, but some allocators are smart: they can issue a madvise(..., MADV_DONTNEED) call which tells the OS "this range doesn't have anything useful, you don't have to preserve its contents in swap". It still leaves the virtual address space mapped in the process and usable later (zero filled) and so it's more efficient than munmap and a subsequent mmap, but it avoid pointlessly swapping freed memory regions to swap.²

The Demonstrated Code

As pointed out in this answer your test with vector<int> isn't really testing anything because an empty, unused std::vector<int> v won't even create the vector object as long as you are using some minimal level of optimization. Even without optimization, no allocation is likely to occur because most vector implementations allocate on first insertion, and not in the constructor. Finally, even if you are using some unusual compiler or library that does an allocation, it will be for a handful of bytes, not the ~72,000 bytes Valgrind is reporting.

You should do something like this to actually see the impact of a vector allocation:

#include <vector>

volatile vector<int> *sink;

int main() {
    std::vector<int> v(12345678);
    sink = &v;
}

That results in actual allocation and de-allocation. It isn't going to change the Valgrind output, however, since the vector allocation is correctly freed before the program exits, so there is no issue as far as Valgrind is concerned.

At a high level, Valgrind basically categorizes things into "definite leaks" and "not freed at exit". The former occur when the program no longer has a reference to a pointer to memory that it allocated. It cannot free such memory and so has leaked it. Memory which hasn't been freed at exit may be a "leak" - i.e., objects that should have been freed, but it may also simply be memory that the developer knew would live the length of the program and so doesn't need to be explicitly freed (because of order-of-destruction issues for globals, especially when shared libraries are involved, it may be very hard to reliably free memory associated with global or static objects even if you wanted to).

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¹ In some cases some deliberately special allocations may outlive the process, such as shared memory and memory mapped files, but that doesn't relate to plain C++ allocations and you can ignore it for the purposes of this discussion.

² Recent Linux kernels also have the Linux-specific MADV_FREE which seems to have similar semantics to MADV_DONTNEED.

1. c++ memory allocator:
2. c++ concepts