Question

我试图了解linux进程内存布局的基本知识，并且得到了以下程序：

#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;
}

启动程序时，/proc/<pid>/maps中的内容为：

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]

这些存储区的用途是什么？

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

然后在运行程序后按Enter键几次。之后显示“线程1中的malloc之前”。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“在线程1中的malloc之后和free之前”之后，它将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？

Answer 1

您的问题涵盖了许多完全不同的内容，因此答案将很长。

第一个问题是

的含义

7fcc37491000-7fcc37691000 ---p 001ba000 08:01 1053757                    /lib/x86_64-linux-gnu/libc-2.19.so

在

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

该不可访问的内存区域是库的相邻ELF段之间的间隙（应该占用连续的内存块）。 ---p保护模式禁止将此间隙用于偶尔的内存分配。如果您在strace(1)加载库的过程中看到以下内容：

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，但为整个库保留空间。这样做是为了允许内核自行选择库的位置。为了保护段mprotect(..., PROT_NONE)之间的任何可能的间隙，请调用；然后使用mmap()将所有剩余的段映射到内存中-这也将适当的内存页的保护模式从---p更改为段所需的任何模式。看看how it actually works，您可能会觉得很有趣。如果您想验证在加载过程中---p间隙是如何形成的，还可以对库的二进制文件使用readelf(1)并使用段的位置和对齐方式进行一些十六进制数学运算，将结果与strace。

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，而后者又是默认堆栈大小pthread），第一个是堆栈保护页面。模式为---p的页面可防止堆栈溢出，并在发生溢出时（由于对此受写保护的页面进行写操作）而触发段错误。

注意：在较早的Linux内核中，线程栈在[stack:TID]文件中用maps名称进行注释，但是此功能已删除，因此我不能保证该映射实际上是一个线程堆栈（尽管看起来像）。但是，您可以使用strace从系统调用child_stack的{{1}}参数中找到确切的线程堆栈位置，并与该映射进行比较。

继续。 第三个问题是

clone()

好吧，这就是7fcc30000000-7fcc30021000 rw-p 00000000 00:00 0 7fcc30021000-7fcc34000000 ---p 00000000 00:00 0中的malloc()分配您所请求的内存的过程。简而言之，整个区域thread 1是一个堆，从中进行分配。从{堆中分配的7fcc30000000-7fcc34000000间隔rw-p将随着您使用7fcc30000000-7fcc30021000请求越来越多的内存而增大。当该堆耗尽时，将使用malloc()请求新的堆。

您可能已经注意到，对于您的问题中的以下映射，我没有任何解释：

mmap()

我无法很快认出这些家伙，并且不确定这些是普通的分配。可能需要单独调查，因为该主题已经太长了。

了解Linux中的进程内存映射

1 个答案: