我试图理解linux进程内存布局的基础,我得到了以下程序:
#include <stdio.h> // standard io
#include <stdlib.h> // C standard library
#include <pthread.h> // threading
#include <unistd.h> // unix standard library
#include <sys/types.h> // system types for linux
// getchar basically is like "read"
// it prompts the user for input
// in this case, the input is thrown away
// which makes similar to a "pause" continuation primitive
// but a pause that is resolved through user input, which we promptly throw away!
void * thread_func (void * arg) {
printf("Before malloc in thread 1\n");
getchar();
char * addr = (char *) malloc(1000);
printf("After malloc and before free in thread 1\n");
getchar();
free(addr);
printf("After free in thread 1\n");
getchar();
}
int main () {
char * addr;
printf("Welcome to per thread arena example::%d\n", getpid());
printf("Before malloc in the main thread\n");
getchar();
addr = (char *) malloc(1000);
printf("After malloc and before free in main thread\n");
getchar();
free(addr);
printf("After free in main thread\n");
getchar();
// pointer to the thread 1
pthread_t thread_1;
// pthread_* functions return 0 upon succeeding, and other numbers upon failing
int pthread_status;
pthread_status = pthread_create(&thread_1, NULL, thread_func, NULL);
if (pthread_status != 0) {
printf("Thread creation error\n");
return -1;
}
// returned status code from thread_1
void * thread_1_status;
pthread_status = pthread_join(thread_1, &thread_1_status);
if (pthread_status != 0) {
printf("Thread join error\n");
return -1;
}
return 0;
}
当我启动程序时,/proc/<pid>/maps
中的内容是:
00400000-00401000 r-xp 00000000 08:01 1323314 /home/oscp/xg/c/memory_layout/a.out
00600000-00601000 r--p 00000000 08:01 1323314 /home/oscp/xg/c/memory_layout/a.out
00601000-00602000 rw-p 00001000 08:01 1323314 /home/oscp/xg/c/memory_layout/a.out
7fcc372d7000-7fcc37491000 r-xp 00000000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
7fcc37491000-7fcc37691000 ---p 001ba000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
7fcc37691000-7fcc37695000 r--p 001ba000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
7fcc37695000-7fcc37697000 rw-p 001be000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
7fcc37697000-7fcc3769c000 rw-p 00000000 00:00 0
7fcc3769c000-7fcc376b5000 r-xp 00000000 08:01 1053877 /lib/x86_64-linux-gnu/libpthread-2.19.so
7fcc376b5000-7fcc378b4000 ---p 00019000 08:01 1053877 /lib/x86_64-linux-gnu/libpthread-2.19.so
7fcc378b4000-7fcc378b5000 r--p 00018000 08:01 1053877 /lib/x86_64-linux-gnu/libpthread-2.19.so
7fcc378b5000-7fcc378b6000 rw-p 00019000 08:01 1053877 /lib/x86_64-linux-gnu/libpthread-2.19.so
7fcc378b6000-7fcc378ba000 rw-p 00000000 00:00 0
7fcc378ba000-7fcc378dd000 r-xp 00000000 08:01 1053733 /lib/x86_64-linux-gnu/ld-2.19.so
7fcc37abe000-7fcc37ac1000 rw-p 00000000 00:00 0
7fcc37ad8000-7fcc37adc000 rw-p 00000000 00:00 0
7fcc37adc000-7fcc37add000 r--p 00022000 08:01 1053733 /lib/x86_64-linux-gnu/ld-2.19.so
7fcc37add000-7fcc37ade000 rw-p 00023000 08:01 1053733 /lib/x86_64-linux-gnu/ld-2.19.so
7fcc37ade000-7fcc37adf000 rw-p 00000000 00:00 0
7ffdc1cff000-7ffdc1d20000 rw-p 00000000 00:00 0 [stack]
7ffdc1dd8000-7ffdc1ddb000 r--p 00000000 00:00 0 [vvar]
7ffdc1ddb000-7ffdc1ddd000 r-xp 00000000 00:00 0 [vdso]
ffffffffff600000-ffffffffff601000 r-xp 00000000 00:00 0 [vsyscall]
这些记忆区域的用途是什么?
7fcc37491000-7fcc37691000 ---p 001ba000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
...
7fcc37abe000-7fcc37ac1000 rw-p 00000000 00:00 0
7fcc37ad8000-7fcc37adc000 rw-p 00000000 00:00 0
运行程序后,我按enter键几次。在打印“在线程1中的malloc之前”之后。内存布局如下所示:
00400000-00401000 r-xp 00000000 08:01 1323314 /home/oscp/xg/c/memory_layout/a.out
00600000-00601000 r--p 00000000 08:01 1323314 /home/oscp/xg/c/memory_layout/a.out
00601000-00602000 rw-p 00001000 08:01 1323314 /home/oscp/xg/c/memory_layout/a.out
00632000-00653000 rw-p 00000000 00:00 0 [heap]
7fcc36ad6000-7fcc36ad7000 ---p 00000000 00:00 0
7fcc36ad7000-7fcc372d7000 rw-p 00000000 00:00 0
7fcc372d7000-7fcc37491000 r-xp 00000000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
7fcc37491000-7fcc37691000 ---p 001ba000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
7fcc37691000-7fcc37695000 r--p 001ba000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
7fcc37695000-7fcc37697000 rw-p 001be000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
7fcc37697000-7fcc3769c000 rw-p 00000000 00:00 0
7fcc3769c000-7fcc376b5000 r-xp 00000000 08:01 1053877 /lib/x86_64-linux-gnu/libpthread-2.19.so
7fcc376b5000-7fcc378b4000 ---p 00019000 08:01 1053877 /lib/x86_64-linux-gnu/libpthread-2.19.so
7fcc378b4000-7fcc378b5000 r--p 00018000 08:01 1053877 /lib/x86_64-linux-gnu/libpthread-2.19.so
7fcc378b5000-7fcc378b6000 rw-p 00019000 08:01 1053877 /lib/x86_64-linux-gnu/libpthread-2.19.so
7fcc378b6000-7fcc378ba000 rw-p 00000000 00:00 0
7fcc378ba000-7fcc378dd000 r-xp 00000000 08:01 1053733 /lib/x86_64-linux-gnu/ld-2.19.so
7fcc37abe000-7fcc37ac1000 rw-p 00000000 00:00 0
7fcc37ad8000-7fcc37adc000 rw-p 00000000 00:00 0
7fcc37adc000-7fcc37add000 r--p 00022000 08:01 1053733 /lib/x86_64-linux-gnu/ld-2.19.so
7fcc37add000-7fcc37ade000 rw-p 00023000 08:01 1053733 /lib/x86_64-linux-gnu/ld-2.19.so
7fcc37ade000-7fcc37adf000 rw-p 00000000 00:00 0
7ffdc1cff000-7ffdc1d20000 rw-p 00000000 00:00 0 [stack]
7ffdc1dd8000-7ffdc1ddb000 r--p 00000000 00:00 0 [vvar]
7ffdc1ddb000-7ffdc1ddd000 r-xp 00000000 00:00 0 [vdso]
ffffffffff600000-ffffffffff601000 r-xp 00000000 00:00 0 [vsyscall]
这两个地区的目的是什么?
7fcc36ad6000-7fcc36ad7000 ---p 00000000 00:00 0
7fcc36ad7000-7fcc372d7000 rw-p 00000000 00:00 0
在打印“malloc之后,在线程1中空闲之前”之后,它会在下面创建另外两个区域:
7fcc30000000-7fcc30021000 rw-p 00000000 00:00 0
7fcc30021000-7fcc34000000 ---p 00000000 00:00 0
这两个地区的目的是什么?
发布于 2018-11-25 03:38:06
你的问题涵盖了许多完全不同的事情,所以答案会很长。
的第一个问题是
7fcc37491000-7fcc37691000 ---p 001ba000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
在……里面
7fcc372d7000-7fcc37491000 r-xp 00000000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
7fcc37491000-7fcc37691000 ---p 001ba000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
7fcc37691000-7fcc37695000 r--p 001ba000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
7fcc37695000-7fcc37697000 rw-p 001be000 08:01 1053757 /lib/x86_64-linux-gnu/libc-2.19.so
这个不可访问的内存区域是库中相邻的ELF段之间的一个空白(它应该占用内存的连续块)。---p
保护模式禁止在偶尔分配内存时使用此间隙。如果在进程加载库时对其进行strace(1)
,您可能会看到如下所示:
mmap(NULL, 1848896, PROT_READ, MAP_PRIVATE|MAP_DENYWRITE, 3</usr/lib/libc-2.28.so>, 0) = 0x7f9673d8f000
mprotect(0x7f9673db1000, 1671168, PROT_NONE) = 0
mmap(0x7f9673db1000, 1355776, PROT_READ|PROT_EXEC, MAP_PRIVATE|MAP_FIXED|MAP_DENYWRITE, 3</usr/lib/libc-2.28.so>, 0x22000) = 0x7f9673db1000
mmap(0x7f9673efc000, 311296, PROT_READ, MAP_PRIVATE|MAP_FIXED|MAP_DENYWRITE, 3</usr/lib/libc-2.28.so>, 0x16d000) = 0x7f9673efc000
mmap(0x7f9673f49000, 24576, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_FIXED|MAP_DENYWRITE, 3</usr/lib/libc-2.28.so>, 0x1b9000) = 0x7f9673f49000
第一个mmap()
将第一个ELF段映射到内存中,但保留了整个库的空间。这样做是为了允许内核酌情选择库的位置。为了保护段之间的任何可能的空白,调用mprotect(..., PROT_NONE)
;然后使用mmap()
将所有剩余的段映射到内存中--这也会将适当内存页的保护模式从---p
更改为该段所需的任何模式。您可以通过查看它实际上是如何工作的来获得一些乐趣。如果要验证加载过程中如何形成---p
间隙,还可以在库的二进制文件中使用readelf(1)
,并对段的位置和对齐进行十六进制运算,并将结果与strace
的输出进行比较。
第二个问题是以下匿名映射:
7fcc36ad6000-7fcc36ad7000 ---p 00000000 00:00 0
7fcc36ad7000-7fcc372d7000 rw-p 00000000 00:00 0
这看起来像thread 1
的线程堆栈。第二个映射是堆栈本身(372d7000
- 36ad7000
= 800000
=8 MiB,这是许多发行版中的默认堆栈大小限制,这反过来又是pthread
的默认堆栈大小),第一个映射是堆栈保护页。这个具有模式---p
的页面保护堆栈不溢出,并在溢出发生时触发分段错误(因为对此写保护的页面进行了写入)。
注意:在旧的Linux内核中,线程堆栈在[stack:TID]
文件中使用maps
名称进行注释,但是这个特性被删除了,所以我不能保证这个映射实际上是一个线程堆栈(尽管它看起来像)。但是,您可以使用strace
从clone()
syscall的child_stack
参数中找到确切的线程堆栈位置,并与此映射进行比较。
还在继续。第三个问题是
7fcc30000000-7fcc30021000 rw-p 00000000 00:00 0
7fcc30021000-7fcc34000000 ---p 00000000 00:00 0
这就是malloc()
in thread 1
为分配所请求的内存所做的事情。简而言之,整个区域7fcc30000000-7fcc34000000
是一个堆,从堆中进行分配。从这个堆中分配的rw-p
间隔7fcc30000000-7fcc30021000
将增长,因为您将使用malloc()
请求越来越多的内存。当这个堆耗尽时,将使用mmap()
请求新的堆。
正如您可能注意到的,对于您的问题中的下列映射,我没有一个解释:
7fcc37abe000-7fcc37ac1000 rw-p 00000000 00:00 0
7fcc37ad8000-7fcc37adc000 rw-p 00000000 00:00 0
我不能很快认出那些家伙,也不确定这些是普通的分配。这可能需要单独调查,因为这个话题已经太长了。
https://stackoverflow.com/questions/53455603
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