freeing "copy-on-write" memory that wasn't changed

Question

I get the idea behind copy-on-write. When I fork, the heap is marked as CoW, and when any process tries to change it, a copy is made. The question is: do I have to free it in a child's process nonetheless? Suppose a parent has a dynamic char *array, then it forks. A child process prints some const char, and exits. The child process did not alter the heap at all. Will there be a memory leak?

edit: My child process prints the array on heap, but doesn't modify it. Valgrind says there is a leak if I do not free that array. No leak/memory errors when I free it.

Berkelium

Answer 1

CoW is just a lazy optimization. You may freely think that fork() always makes the full copy of the process (in terms of memory at least) without any delay. But…

If you did prepare dynamic data chunk to "pass" to fork's child, then after fork you have two processes with two dynamic data chunks: parent and child (both are copies). When child exits, it's copy of memory is reclaimed, but parent should free that chunk right after fork by itself.

To be more clear, here is an example:

char *buf = malloc(123456);
// … fill buf for child …

int res = fork();

if (res == -1) {
    fprintf(stderr, "fork failed\n");
    exit(EXIT_FAILURE);
}

if (res == 0) {
    // this is child process
    // … do work with buf …
    _Exit(EXIT_SUCCESS); // child reclaims buf by means of exit
}

// this is parent process
free(buf); // we don't need it in parent

// … other parent tasks here …

CoW is also very useful optimization in fork-exec technique, where child does nothing but exec with prepared arguments. exec replaces current process with specified executable image, retaining open descriptors and other things (more in man 2 execve). The only page that is copied after such fork is only current stack frame, making fork-exec very effective.

Some systems also provide vfork, that is very restrictive unfair version of fork, but on systems without CoW that is the only way to vfork-exec efficiently.

Dionnadionne

So I have to free it even if CoW is implemented and the memory wasn't really copied (not altered by a child)? – Berkelium

@Krzysiek: Yes you have to free it. Once the original program exits there is one copy of the memory left in the forked program and that needs to be freed. If you didn't free it, the forked program would just consume more and more memory. All memory will be freed on program exit of course. – Castigate

@Berkelium CoW is a result of [theoretical] observation that child processes do change only some of the pages, but not all. So, delayed copy is a great benefit in that some of the pages will not be copied at all. This is completely invisible kernel-side optimization and has nothing to do with regular memory management. Remember: if you allocate memory, you are responsible to free it, with free or exit. If you fork, responsibilities also double. – Dionnadionne

@Berkelium Added an example. – Dionnadionne

Answer 2

First the logical (process centered) view:

When you fork a process, the entire address space is copied into a new process as is. Your heap is essentially duplicated in both processes, and both processes can continue to use it just like one process could if fork() had never been called. Both processes can free an allocation that was done before the fork(), and they must do so if they want to reuse the address range connected to the allocation. CoW mappings are only an optimization that does not change these semantics.

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Now the physical (system centered) view:

Your system kernel does not know about data ranges you have allocated using malloc(), it only knows about the memory pages it has allocated to the process at the request of malloc(). When you call fork() it marks all these pages as CoW, and references them from both processes. If any of the two processes writes to any of the CoW pages while the other process still exists, it will trap into the system which copies the entire page. And if one of the processes exits, it will at the very least lower the reference count of these pages, so that they do not have to be copied anymore.

So, what happens when you call free() in the child before exiting?
Well, the free() function will most likely write to the page containing the memory allocation to tell malloc() that the block is available again. This will trap into the system and copy the page, expect this operation to take a microsecond or two. If your parent process calls free() while the child is still alive, the same will happen. However, if your child does not free the page and exits, the kernel will know that it does not have to perform CoW anymore. If the parent frees and reuses the memory region afterwards, no copy needs to be done.

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I assume, that what your child does is simply to check for some error condition, and exit immediately if it is met. In that case, the most prudent approach is to forget about calling free() in the child, and let the system do its work.

Histopathology