What does the brk() system call do? - c

According to Linux programmers manual:
brk() and sbrk() change the location of the program break, which
defines the end of the process's data segment.
What does the data segment mean over here? Is it just the data segment or data, BSS, and heap combined?
According to wiki Data segment:
Sometimes the data, BSS, and heap areas are collectively referred to as the "data segment".
I see no reason for changing the size of just the data segment. If it is data, BSS and heap collectively then it makes sense as heap will get more space.
Which brings me to my second question. In all the articles I read so far, author says that heap grows upward and stack grows downward. But what they do not explain is what happens when heap occupies all the space between heap and stack?

In the diagram you posted, the "break"—the address manipulated by brk and sbrk—is the dotted line at the top of the heap.
The documentation you've read describes this as the end of the "data segment" because in traditional (pre-shared-libraries, pre-mmap) Unix the data segment was continuous with the heap; before program start, the kernel would load the "text" and "data" blocks into RAM starting at address zero (actually a little above address zero, so that the NULL pointer genuinely didn't point to anything) and set the break address to the end of the data segment. The first call to malloc would then use sbrk to move the break up and create the heap in between the top of the data segment and the new, higher break address, as shown in the diagram, and subsequent use of malloc would use it to make the heap bigger as necessary.
Meantime, the stack starts at the top of memory and grows down. The stack doesn't need explicit system calls to make it bigger; either it starts off with as much RAM allocated to it as it can ever have (this was the traditional approach) or there is a region of reserved addresses below the stack, to which the kernel automatically allocates RAM when it notices an attempt to write there (this is the modern approach). Either way, there may or may not be a "guard" region at the bottom of the address space that can be used for stack. If this region exists (all modern systems do this) it is permanently unmapped; if either the stack or the heap tries to grow into it, you get a segmentation fault. Traditionally, though, the kernel made no attempt to enforce a boundary; the stack could grow into the heap, or the heap could grow into the stack, and either way they would scribble over each other's data and the program would crash. If you were very lucky it would crash immediately.
I'm not sure where the number 512GB in this diagram comes from. It implies a 64-bit virtual address space, which is inconsistent with the very simple memory map you have there. A real 64-bit address space looks more like this:
Legend: t: text, d: data, b: BSS
This is not remotely to scale, and it shouldn't be interpreted as exactly how any given OS does stuff (after I drew it I discovered that Linux actually puts the executable much closer to address zero than I thought it did, and the shared libraries at surprisingly high addresses). The black regions of this diagram are unmapped -- any access causes an immediate segfault -- and they are gigantic relative to the gray areas. The light-gray regions are the program and its shared libraries (there can be dozens of shared libraries); each has an independent text and data segment (and "bss" segment, which also contains global data but is initialized to all-bits-zero rather than taking up space in the executable or library on disk). The heap is no longer necessarily continous with the executable's data segment -- I drew it that way, but it looks like Linux, at least, doesn't do that. The stack is no longer pegged to the top of the virtual address space, and the distance between the heap and the stack is so enormous that you don't have to worry about crossing it.
The break is still the upper limit of the heap. However, what I didn't show is that there could be dozens of independent allocations of memory off there in the black somewhere, made with mmap instead of brk. (The OS will try to keep these far away from the brk area so they don't collide.)

Minimal runnable example
What does brk( ) system call do?
Asks the kernel to let you you read and write to a contiguous chunk of memory called the heap.
If you don't ask, it might segfault you.
Without brk:
#define _GNU_SOURCE
#include <unistd.h>
int main(void) {
/* Get the first address beyond the end of the heap. */
void *b = sbrk(0);
int *p = (int *)b;
/* May segfault because it is outside of the heap. */
*p = 1;
return 0;
}
With brk:
#define _GNU_SOURCE
#include <assert.h>
#include <unistd.h>
int main(void) {
void *b = sbrk(0);
int *p = (int *)b;
/* Move it 2 ints forward */
brk(p + 2);
/* Use the ints. */
*p = 1;
*(p + 1) = 2;
assert(*p == 1);
assert(*(p + 1) == 2);
/* Deallocate back. */
brk(b);
return 0;
}
GitHub upstream.
The above might not hit a new page and not segfault even without the brk, so here is a more aggressive version that allocates 16MiB and is very likely to segfault without the brk:
#define _GNU_SOURCE
#include <assert.h>
#include <unistd.h>
int main(void) {
void *b;
char *p, *end;
b = sbrk(0);
p = (char *)b;
end = p + 0x1000000;
brk(end);
while (p < end) {
*(p++) = 1;
}
brk(b);
return 0;
}
Tested on Ubuntu 18.04.
Virtual address space visualization
Before brk:
+------+ <-- Heap Start == Heap End
After brk(p + 2):
+------+ <-- Heap Start + 2 * sizof(int) == Heap End
| |
| You can now write your ints
| in this memory area.
| |
+------+ <-- Heap Start
After brk(b):
+------+ <-- Heap Start == Heap End
To better understand address spaces, you should make yourself familiar with paging: How does x86 paging work?.
Why do we need both brk and sbrk?
brk could of course be implemented with sbrk + offset calculations, both exist just for convenience.
In the backend, the Linux kernel v5.0 has a single system call brk that is used to implement both: https://github.com/torvalds/linux/blob/v5.0/arch/x86/entry/syscalls/syscall_64.tbl#L23
12 common brk __x64_sys_brk
Is brk POSIX?
brk used to be POSIX, but it was removed in POSIX 2001, thus the need for _GNU_SOURCE to access the glibc wrapper.
The removal is likely due to the introduction mmap, which is a superset that allows multiple range to be allocated and more allocation options.
I think there is no valid case where you should to use brk instead of malloc or mmap nowadays.
brk vs malloc
brk is one old possibility of implementing malloc.
mmap is the newer stricly more powerful mechanism which likely all POSIX systems currently use to implement malloc. Here is a minimal runnable mmap memory allocation example.
Can I mix brk and malloc?
If your malloc is implemented with brk, I have no idea how that can possibly not blow up things, since brk only manages a single range of memory.
I could not however find anything about it on the glibc docs, e.g.:
https://www.gnu.org/software/libc/manual/html_mono/libc.html#Resizing-the-Data-Segment
Things will likely just work there I suppose since mmap is likely used for malloc.
See also:
What's unsafe/legacy about brk/sbrk?
Why does calling sbrk(0) twice give a different value?
More info
Internally, the kernel decides if the process can have that much memory, and earmarks memory pages for that usage.
This explains how the stack compares to the heap: What is the function of the push / pop instructions used on registers in x86 assembly?

You can use brk and sbrk yourself to avoid the "malloc overhead" everyone's always complaining about. But you can't easily use this method in conjuction with malloc so it's only appropriate when you don't have to free anything. Because you can't. Also, you should avoid any library calls which may use malloc internally. Ie. strlen is probably safe, but fopen probably isn't.
Call sbrk just like you would call malloc. It returns a pointer to the current break and increments the break by that amount.
void *myallocate(int n){
return sbrk(n);
}
While you can't free individual allocations (because there's no malloc-overhead, remember), you can free the entire space by calling brk with the value returned by the first call to sbrk, thus rewinding the brk.
void *memorypool;
void initmemorypool(void){
memorypool = sbrk(0);
}
void resetmemorypool(void){
brk(memorypool);
}
You could even stack these regions, discarding the most recent region by rewinding the break to the region's start.
One more thing ...
sbrk is also useful in code golf because it's 2 characters shorter than malloc.

There is a special designated anonymous private memory mapping (traditionally located just beyond the data/bss, but modern Linux will actually adjust the location with ASLR). In principle it's no better than any other mapping you could create with mmap, but Linux has some optimizations that make it possible to expand the end of this mapping (using the brk syscall) upwards with reduced locking cost relative to what mmap or mremap would incur. This makes it attractive for malloc implementations to use when implementing the main heap.

malloc uses brk system call to allocate memory.
include
int main(void){
char *a = malloc(10);
return 0;
}
run this simple program with strace, it will call brk system.

I can answer your second question. Malloc will fail and return a null pointer. That's why you always check for a null pointer when dynamically allocating memory.

The heap is placed last in the program's data segment. brk() is used to change (expand) the size of the heap. When the heap cannot grow any more any malloc call will fail.

The data segment is the portion of memory that holds all your static data, read in from the executable at launch and usually zero-filled.

Related

Will malloc round up to the nearest page size?

I'm not sure if I'm asking a noob question here, but here I go. I also searched a lot for a similar question, but I got nothing.
So, I know how mmap and brk work and that, regardless of the length you enter, it will round it up to the nearest page boundary. I also know malloc uses brk/sbrk or mmap (At least on Linux/Unix systems) but this raises the question: does malloc also round up to the nearest page size? For me, a page size is 4096 bytes, so if I want to allocate 16 bytes with malloc, 4096 bytes is... a lot more than I asked for.
The basic job of malloc and friends is to manage the fact that the OS can generally only (efficiently) deal with large allocations (whole pages and extents of pages), while programs often need smaller chunks and finer-grained management.
So what malloc (generally) does, is that the first time it is called, it allocates a larger amount of memory from the system (via mmap or sbrk -- maybe one page or maybe many pages), and uses a small amount of that for some data structures to track the heap use (where the heap is, what parts are in use and what parts are free) and then marks the rest of that space as free. It then allocates the memory you requested from that free space and keeps the rest available for subsequent malloc calls.
So the first time you call malloc for eg 16 bytes, it will uses mmap or sbrk to allocate a large chunk (maybe 4K or maybe 64K or maybe 16MB or even more) and initialize that as mostly free and return you a pointer to 16 bytes somewhere. A second call to malloc for another 16 bytes will just return you another 16 bytes from that pool -- no need to go back to the OS for more.
As your program goes ahead mallocing more memory it will just come from this pool, and free calls will return memory to the free pool. If it generally allocates more than it frees, eventually that free pool will run out, and at that point, malloc will call the system (mmap or sbrk) to get more memory to add to the free pool.
This is why if you monitor a process that is allocating and freeing memory with malloc/free with some sort of process monitor, you will generally just see the memory use go up (as the free pool runs out and more memory is requested from the system), and generally will not see it go down -- even though memory is being freed, it generally just goes back to the free pool and is not unmapped or returned to the system. There are some exceptions -- particularly if very large blocks are involved -- but generally you can't rely on any memory being returned to the system until the process exits.
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
#include <unistd.h>
int main(void) {
void *a = malloc(1);
void *b = malloc(1);
uintptr_t ua = (uintptr_t)a;
uintptr_t ub = (uintptr_t)b;
size_t page_size = getpagesize();
printf("page size: %zu\n", page_size);
printf("difference: %zd\n", (ssize_t)(ub - ua));
printf("offsets from start of page: %zu, %zu\n",
(size_t)ua % page_size, (size_t)ub % page_size);
}
prints
page_size: 4096
difference: 32
offsets from start of page: 672, 704
So clearly it is not rounded to page size in this case, which proves that it is not always rounded to page size.
It will hit mmap if you change allocation to some arbitrary large size. For example:
void *a = malloc(10000001);
void *b = malloc(10000003);
and I get:
page size: 4096
difference: -10002432
offsets from start of page: 16, 16
And clearly the starting address is still not page aligned; the bookkeeping must be stored below the pointer and the pointer needs to be sufficiently aligned for the largest alignment generally needed - you can reason this with free - if free is just given a pointer but it needs to figure out the size of the allocation, where could it look for it, and only two choices are feasible: in a separate data structure that lists all base pointers and their allocation sizes, or at some offset below the current pointer. And only one of them is sane.

Does malloc only assign space in linear region structure vm_area_t insteading of real memory

I read in "Understanding the linux kernel" that when malloc is invokded in user space, the kernel only add an linear region in the vm_area_t structure, insteading of allocating space in memory, which is called ostponing the allocation, and this space allocated in linear region can only be used when page interrupts occur. But if no page can be assigned during the page interrupts, isn't the user cheated when it invokes a malloc?
The malloc() call wrapped by glibc will first allocate from the heap memory maintained by glibc. (see what happens in the kernel during malloc?)
If glibc runs out heap memory, sys_brk is invoked to allocate more.
The kernel only allocate the memory to user request when it first touches the allocate memory (mostly page fault handler).
Thus, I guess calloc will returned a real allocated one because it initialized the page to zero (from man page), which touches the page by scrubbing its content.
If the new page can't be allocated in the fault handler, which means the system has very low memory, either in heavy swap or is going to OOM.
One small thing is that Linux always scrub the memory when it allocates to user process, because it doesn't do that when the page is returned to buddy allocator.
Memory allocated by malloc is not in real memory before it is written to. calloc on the other hand allocate real memory at ones.
From man the man page:
By default, Linux follows an optimistic memory allocation strategy.
This means that when malloc() returns non-NULL there is no guarantee
that the memory really is available. This is a really bad bug. In
case it turns out that the system is out of memory, one or more pro‐
cesses will be killed by the infamous OOM killer. In case Linux is
employed under circumstances where it would be less desirable to sud‐
denly lose some randomly picked processes, and moreover the kernel
version is sufficiently recent, one can switch off this overcommitting
behavior using a command like:
# echo 2 > /proc/sys/vm/overcommit_memory
looking around in source of glibc and found this for what I believe is calloc:
if (__builtin_expect (hook != NULL, 0)) {
sz = bytes;
mem = (*hook)(sz, RETURN_ADDRESS (0));
if(mem == 0)
return 0;
#ifdef HAVE_MEMCPY
return memset(mem, 0, sz);
#else
while(sz > 0) ((char*)mem)[--sz] = 0; /* rather inefficient */
return mem;
#endif
}

How do I calculate beforehand how much memory calloc would allocate?

I basically have this piece of code.
char (* text)[1][80];
text = calloc(2821522,80);
The way I calculated it, that calloc should have allocated 215.265045 megabytes of RAM, however, the program in the end exceeded that number and allocated nearly 700mb of ram.
So it appears I cannot properly know how much memory that function will allocate.
How does one calculate that propery?
calloc (and malloc for that matter) is free to allocate as much space as it needs to satisfy the request.
So, no, you cannot tell in advance how much it will actually give you, you can only assume that it's given you the amount you asked for.
Having said that, 700M seems a little excessive so I'd be investigating whether the calloc was solely responsible for that by, for example, a program that only does the calloc and nothing more.
You might also want to investigate how you're measuring that memory usage.
For example, the following program:
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
int main (void) {
char (* text)[1][80];
struct mallinfo mi;
mi = mallinfo(); printf ("%d\n", mi.uordblks);
text = calloc(2821522,80);
mi = mallinfo(); printf ("%d\n", mi.uordblks);
return 0;
}
outputs, on my system:
66144
225903256
meaning that the calloc has allocated 225,837,112 bytes which is only a smidgeon (115,352 bytes or 0.05%) above the requested 225,721,760.
Well it depends on the underlying implementation of malloc/calloc.
It generally works like this - there's this thing called the heap pointer which points to the top of the heap - the area from where dynamic memory gets allocated. When memory is first allocated, malloc internally requests x amount of memory from the kernel - i.e. the heap pointer increments by a certain amount to make that space available. That x may or may not be equal to the size of the memory block you requested (it might be larger to account for future mallocs). If it isn't, then you're given at least the amount of memory you requested(sometimes you're given more memory because of alignment issues). The rest is made part of an internal free list maintained by malloc. To sum it up malloc has some underlying data structures and a lot depends on how they are implemented.
My guess is that the x amount of memory was larger (for whatever reason) than you requested and hence malloc/calloc was holding on to the rest in its free list. Try allocating some more memory and see if the footprint increases.

Problem usage memory in C

Please help :)
OS : Linux
Where in " sleep(1000);", at this time "top (display Linux tasks)" wrote me 7.7 %MEM use.
valgrind : not found memory leak.
I understand, wrote correctly and all malloc result is NULL.
But Why in this time "sleep" my program NOT decreased memory ? What missing ?
Sorry for my bad english, Thanks
~ # tmp_soft
For : Is it free?? no
Is it free?? yes
For 0
For : Is it free?? no
Is it free?? yes
For 1
END : Is it free?? yes
END
~ #top
PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND
23060 root 20 0 155m 153m 448 S 0 7.7 0:01.07 tmp_soft
Full source : tmp_soft.c
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <unistd.h>
struct cache_db_s
{
int table_update;
struct cache_db_s * p_next;
};
void free_cache_db (struct cache_db_s ** cache_db)
{
struct cache_db_s * cache_db_t;
while (*cache_db != NULL)
{
cache_db_t = *cache_db;
*cache_db = (*cache_db)->p_next;
free(cache_db_t);
cache_db_t = NULL;
}
printf("Is it free?? %s\n",*cache_db==NULL?"yes":"no");
}
void make_cache_db (struct cache_db_s ** cache_db)
{
struct cache_db_s * cache_db_t = NULL;
int n = 10000000;
for (int i=0; i = n; i++)
{
if ((cache_db_t=malloc(sizeof(struct cache_db_s)))==NULL) {
printf("Error : malloc 1 -> cache_db_s (no free memory) \n");
break;
}
memset(cache_db_t, 0, sizeof(struct cache_db_s));
cache_db_t->table_update = 1; // tmp
cache_db_t->p_next = *cache_db;
*cache_db = cache_db_t;
cache_db_t = NULL;
}
}
int main(int argc, char **argv)
{
struct cache_db_s * cache_db = NULL;
for (int ii=0; ii 2; ii++) {
make_cache_db(&cache_db);
printf("For : Is it free?? %s\n",cache_db==NULL?"yes":"no");
free_cache_db(&cache_db);
printf("For %d \n", ii);
}
printf("END : Is it free?? %s\n",cache_db==NULL?"yes":"no");
printf("END \n");
sleep(1000);
return 0;
}
For good reasons, virtually no memory allocator returns blocks to the OS
Memory can only be removed from your program in units of pages, and even that is unlikely to be observed.
calloc(3) and malloc(3) do interact with the kernel to get memory, if necessary. But very, very few implementations of free(3) ever return memory to the kernel1, they just add it to a free list that calloc() and malloc() will consult later in order to reuse the released blocks. There are good reasons for this design approach.
Even if a free() wanted to return memory to the system, it would need at least one contiguous memory page in order to get the kernel to actually protect the region, so releasing a small block would only lead to a protection change if it was the last small block in a page.
Theory of Operation
So malloc(3) gets memory from the kernel when it needs it, ultimately in units of discrete page multiples. These pages are divided or consolidated as the program requires. Malloc and free cooperate to maintain a directory. They coalesce adjacent free blocks when possible in order to be able to provide large blocks. The directory may or may not involve using the memory in freed blocks to form a linked list. (The alternative is a bit more shared-memory and paging-friendly, and it involves allocating memory specifically for the directory.) Malloc and free have little if any ability to enforce access to individual blocks even when special and optional debugging code is compiled into the program.
1. The fact that very few implementations of free() attempt to return memory to the system is not at all due to the implementors slacking off.Interacting with the kernel is much slower than simply executing library code, and the benefit would be small. Most programs have a steady-state or increasing memory footprint, so the time spent analyzing the heap looking for returnable memory would be completely wasted. Other reasons include the fact that internal fragmentation makes page-aligned blocks unlikely to exist, and it's likely that returning a block would fragment blocks to either side. Finally, the few programs that do return large amounts of memory are likely to bypass malloc() and simply allocate and free pages anyway.
If you're trying to establish whether your program has a memory leak, then top isn't the right tool for the job (valrind is).
top shows memory usage as seen by the OS. Even if you call free, there is no guarantee that the freed memory would get returned to the OS. Typically, it wouldn't. Nonetheless, the memory does become "free" in the sense that your process can use it for subsequent allocations.
edit If your libc supports it, you could try experimenting with M_TRIM_THRESHOLD. Even if you do follow this path, it's going to be tricky (a single used block sitting close to the top of the heap would prevent all free memory below it from being released to the OS).
Generally free() doesn't give back physical memory to OS, they are still mapped in your process's virtual memory. If you allocate a big chunk of memory, libc may allocate it by mmap(); then if you free it, libc may release the memory to OS by munmap(), in this case, top will show that your memory usage comes down.
So, if you want't to release memory to OS explicitly, you can use mmap()/munmap().
When you free() memory, it is returned to the standard C library's pool of memory, and not returned to the operating system. In the vision of the operating system, as you see it through top, the process is still "using" this memory. Within the process, the C library has accounted for the memory and could return the same pointer from malloc() in the future.
I will explain it some more with a different beginning:
During your calls to malloc, the standard library implementation may determine that the process does not have enough allocated memory from the operating system. At that time, the library will make a system call to receive more memory from the operating system to the process (for example, sbrk() or VirtualAlloc() system calls on Unix or Windows, respectively).
After the library requests additional memory from the operating system, it adds this memory to its structure of memory available to return from malloc. Later calls to malloc will use this memory until it runs out. Then, the library asks the operating system for even more memory.
When you free memory, the library usually does not return the memory to the operating system. There are many reasons for this. One reason is that the library author believed you will call malloc again. If you will not call malloc again, your program will probably end soon. Either case, there is not much advantage to return the memory to the operating system.
Another reason that the library may not return the memory to the operating system is that the memory from operating system is allocated in large, contiguous ranges. It could only be returned when an entire contiguous range is no longer in use. The pattern of calling malloc and free may not clear the entire range of use.
Two problems:
In make_cache_db(), the line
for (int i=0; i = n; i++)
should probably read
for (int i=0; i<n; i++)
Otherwise, you'll only allocate a single cache_db_s node.
The way you're assigning cache_db in make_cache_db() seems to be buggy. It seems that your intention is to return a pointer to the first element of the linked list; but because you're reassigning cache_db in every iteration of the loop, you'll end up returning a pointer to the last element of the list.
If you later free the list using free_cache_db(), this will cause you to leak memory. At the moment, though, this problem is masked by the bug described in the previous bullet point, which causes you to allocate lists of only length 1.
Independent of these bugs, the point raised by aix is very valid: The runtime library need not return all free()d memory to the operating system.

Heap size limitation in C

I have a doubt regarding heap in program execution layout diagram of a C program.
I know that all the dynamically allocated memory is allotted in heap which grows dynamically. But I would like to know what is the max heap size for a C program ??
I am just attaching a sample C program ... here I am trying to allocate 1GB memory to string and even doing the memset ...
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
char *temp;
mybuffer=malloc(1024*1024*1024*1);
temp = memset(mybuffer,0,(1024*1024*1024*1));
if( (mybuffer == temp) && (mybuffer != NULL))
printf("%x - %x\n", mybuffer, &mybuffer[((1024*1024*1024*1)-1)]]);
else
printf("Wrong\n");
sleep(20);
free(mybuffer);
return 0;
}
If I run above program in 3 instances at once then malloc should fail atleast in one instance [I feel so] ... but still malloc is successfull.
If it is successful can I know how the OS takes care of 3GB of dynamically allocated memory.
Your machine is very probably overcomitting on RAM, and not using the memory until you actually write it. Try writing to each block after allocating it, thus forcing the operating system to ensure there's real RAM mapped to the address malloc() returned.
From the linux malloc page,
BUGS
By default, Linux follows an optimistic memory allocation strategy.
This means that when malloc() returns non-NULL there is no guarantee
that the memory really is available. This is a really bad bug. In
case it turns out that the system is out of memory, one or more pro‐
cesses will be killed by the infamous OOM killer. In case Linux is
employed under circumstances where it would be less desirable to sud‐
denly lose some randomly picked processes, and moreover the kernel ver‐
sion is sufficiently recent, one can switch off this overcommitting
behavior using a command like:
# echo 2 > /proc/sys/vm/overcommit_memory
See also the kernel Documentation directory, files vm/overcommit-
accounting and sysctl/vm.txt.
You're mixing up physical memory and virtual memory.
http://apollo.lsc.vsc.edu/metadmin/references/sag/x1752.html
http://en.wikipedia.org/wiki/Virtual_memory
http://duartes.org/gustavo/blog/post/anatomy-of-a-program-in-memory
Malloc will allocate the memory but it does not write to any of it. So if the virtual memory is available then it will succeed. It is only when you write something to it will the real memory need to be paged to the page file.
Calloc if memory serves be correctly(!) write zeros to each byte of the allocated memory before returning so will need to allocate the pages there and then.

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