为什么在预加载到Dash时，共享库的fini函数没有被调用？

huangapple
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英文:

why isn't shared library fini function called when preloaded to dash?

问题

我使用的是最新的Ubuntu Linux。

这是一个包含在加载和卸载时调用的函数的共享库：

shared.c：

  1. #include <fcntl.h>
  2. #include <sys/stat.h>
  4. void init() {
  5. 	open("init", O_RDWR | O_CREAT, S_IRUSR | S_IWUSR);
  6. }
  8. void fini() {
  9. 	open("fini", O_RDWR | O_CREAT, S_IRUSR | S_IWUSR);
  10. }

它与以下命令编译：

gcc -shared -fPIC -Wl,-init=init -Wl,-fini=fini shared.c -o shared.so

然后我执行以下步骤：

  1. $ rm init fini
  2. $ LD_PRELOAD=$PWD/shared.so dash /dev/null
  3. $ echo $?
  4. 0
  5. $ ls init
  6. init
  7. $ ls fini
  8. ls: 无法访问'fini': 没有那个文件或目录

所以... 加载函数被调用，但卸载函数没有被调用。

如果我将dash替换为bash，两者都被调用。

使用__attribute__((destructor))没有任何区别。

为什么对于dash卸载函数没有被调用？

根据Marco Bonelli的请求添加：

  1. $ file $(which dash)
  2. /usr/bin/dash: ELF 64-bit LSB pie 可执行文件, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, BuildID[sha1]=f7ab02fc1b8ff61b41647c1e16ec9d95ba5de9f0, for GNU/Linux 3.2.0, stripped
  4. $ ldd $(which dash)
  5.         linux-vdso.so.1 (0x00007ffd931c0000)
  6.         libc.so.6 => /lib/x86_64-linux-gnu/libc.so.6 (0x00007f6b19e84000)
  7.         /lib64/ld-linux-x86-64.so.2 (0x00007f6b1a0ec000)
英文:

I am on the latest Ubuntu Linux.

Here is a shared library with functions that are called when loading and unloading:

shared.c:

  1. #include &lt;fcntl.h&gt;
  2. #include &lt;sys/stat.h&gt;
  4. void init() {
  5. 	open(&quot;init&quot;, O_RDWR | O_CREAT, S_IRUSR | S_IWUSR);
  6. }
  8. void fini() {
  9. 	open(&quot;fini&quot;, O_RDWR | O_CREAT, S_IRUSR | S_IWUSR);
  10. }

and this is compiled with

gcc -shared -fPIC -Wl,-init=init -Wl,-fini=fini shared.c -o shared.so

and then I do this:

  1. $ rm init fini
  2. $ LD_PRELOAD=$PWD/shared.so  dash /dev/null
  3. $ echo $?
  4. 0
  5. $ ls init
  6. init
  7. $ ls fini
  8. ls: cannot access &#39;fini&#39;: No such file or directory

So... loading function is called, but unloading function is not

If I replace dash with bash , both are called.

Using __attribute__((destructor)) does not make a difference.

Why isn't the unloading function called for dash?

Added per Marco Bonelli's request:

  1. $ file $(which dash)
  2. /usr/bin/dash: ELF 64-bit LSB pie executable, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, BuildID[sha1]=f7ab02fc1b8ff61b41647c1e16ec9d95ba5de9f0, for GNU/Linux 3.2.0, stripped
  4. $ ldd $(which dash)
  5.         linux-vdso.so.1 (0x00007ffd931c0000)
  6.         libc.so.6 =&gt; /lib/x86_64-linux-gnu/libc.so.6 (0x00007f6b19e84000)
  7.         /lib64/ld-linux-x86-64.so.2 (0x00007f6b1a0ec000)

答案1

得分: 2

因为dash在这里调用_exit，它调用exit_group系统调用，这里立即终止程序。

让我们也做一个小例子：

  1. # Makefile
  2. all:
  3.         gcc -shared -fPIC -Wl,-init=init -Wl,-fini=fini -Wl,-soname,shared.so shared.c -o shared.so
  4.         gcc main.c -o main.out
  5.         LD_PRELOAD=$(PWD)/shared.so ./main.out /dev/null
  7. # main.c
  8. #include <unistd.h>
  9. int main() {
  10.         _exit(0);
  11. }
  13. # shared.c
  14. #include <fcntl.h>
  15. #include <sys/stat.h>
  16. #include <unistd.h>
  17. #include <string.h>
  18. void out(const char *str) {
  19.         int fd = open("/dev/tty", O_WRONLY);
  20.         write(fd, str, strlen(str));
  21.         close(fd);
  22. }
  23. void init() {
  24.         out("\nstart\n\n");
  25. }
  26. void fini() {
  27.         out("\nstop\n\n");
  28. }

执行时不会输出"stop"：

  1. $ make
  2. gcc -shared -fPIC -Wl,-init=init -Wl,-fini=fini -Wl,-soname,shared.so shared.c -o shared.so
  3. gcc main.c -o main.out
  4. LD_PRELOAD=..../shared.so ./main.out /dev/null
  6. start
英文:

> why isn't shared library fini function called when preloaded to dash?

Because dash calls _exit here which calls exit_group syscall here which terminates the program immediately.

Let's also do a small example:

  1. # Makefile
  2. all:
  3.         gcc -shared -fPIC -Wl,-init=init -Wl,-fini=fini -Wl,-soname,shared.so shared.c -o shared.so
  4.         gcc main.c -o main.out
  5.         LD_PRELOAD=$(PWD)/shared.so ./main.out /dev/null
  7. # main.c
  8. #include &lt;unistd.h&gt;
  9. int main() {
  10.         _exit(0);
  11. }
  13. # shared.c
  14. #include &lt;fcntl.h&gt;
  15. #include &lt;sys/stat.h&gt;
  16. #include &lt;unistd.h&gt;
  17. #include &lt;string.h&gt;
  18. void out(const char *str) {
  19.         int fd = open(&quot;/dev/tty&quot;, O_WRONLY);
  20.         write(fd, str, strlen(str));
  21.         close(fd);
  22. }
  23. void init() {
  24.         out(&quot;\nstart\n\n&quot;);
  25. }
  26. void fini() {
  27.         out(&quot;\nstop\n\n&quot;);
  28. }

On execution does not output stop:

  1. $ make
  2. gcc -shared -fPIC -Wl,-init=init -Wl,-fini=fini -Wl,-soname,shared.so shared.c -o shared.so
  3. gcc main.c -o main.out
  4. LD_PRELOAD=..../shared.so ./main.out /dev/null
  6. start

答案2

得分: 1

问题不在于“redirection”和“touch”之间的区别，因为在您的“dash/touch”运行中，已加载了shared.so两次，一次用于dash，一次用于touch。
“fini”仅在“touch”中触发。

您可以通过以下两次运行来看到差异：

  1. $ LD_PRELOAD=$PWD/shared.so  dash  /dev/null
  2. # fini未被调用
  3. $ LD_PRELOAD=$PWD/shared.so  touch /dev/null
  4. # fini被调用

因此，这与“dash”有关。

希望这为您的调查提供了更多材料。

英文:

I should have put this as a comment, but I put it as an answer for formatting purposes.

The problem is not about the difference between redirection and touch, because in your dash/touch run, shared.so has been loaded twice, once for dash, once for touch.
fini fired only with touch.

You can see the diffrence with these two runs :

  1. $ LD_PRELOAD=$PWD/shared.so  dash  /dev/null
  2. # fini is not called
  3. $ LD_PRELOAD=$PWD/shared.so  touch /dev/null
  4. # fini is called

So this has something to do with dash.

Hope this provide more materials for your investigation.

huangapple
  • 本文由 发表于 2023年3月7日 14:40:58
  • 转载请务必保留本文链接：https://go.coder-hub.com/75658705.html
  • ld-preload
  • linux
  • linux-kernel
  • shared-libraries
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