What does the brk() system call do?
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:
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 that what happens when heap occupies all the space between heap and stack?
I see a lot of partial answers but no complete answer. Here's that picture you posted again:
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:
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;
}
Tested on Ubuntu 14.04.
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?
More info
brk to be POSIX, but 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.
Internally, the kernel decides if the process can have that much memory, and earmarks memory pages for that usage.
brk and mmap are the common underlying mechanisms that libc uses to implement malloc in POSIX systems.
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 .
下一篇: brk（）系统调用是做什么的？