在Linux中更改进程名称而不更改argv [0]

I need to modify the process name of my program in C language.
I precise, this is not the name of a thread that I want to change.
I want to change the name of my program, but the only solution I found, is to modify the value of argv[0].
I also found another solution with prctl(PR_SET_NAME, "newname"), but this solution doesn't work.

我需要用C语言修改程序的进程名称。我确切地说，这不是我想要改变的线程的名称。我想更改程序的名称，但我找到的唯一解决方案是修改argv [0]的值。我还发现了另一个使用prctl（PR_SET_NAME，“newname”）的解决方案，但此解决方案不起作用。

2 个解决方案

#1

The differences between invoking prctl and modify argv[0] are:

调用prctl和修改argv [0]之间的区别是：

modify argv[0] changes information in /proc/$pid/cmdline
修改argv [0]更改/ proc / $ pid / cmdline中的信息
invoking prctl(PR_SET_NAME) changes information in /proc/$pid/status
调用prctl（PR_SET_NAME）更改/ proc / $ pid / status中的信息

That means you will get difference name of your process issuing ps -a and ps -ax.

这意味着您将获得发出ps -a和ps -ax的进程的不同名称。

If you expects same process name for different arguments while executing ps, you can do them both (i.e., change argv[0] and invoke prctl).

如果在执行ps时期望不同参数的进程名称相同，则可以同时执行它们（即，更改argv [0]并调用prctl）。

Hope the answer helps.

希望答案有所帮助。

#2

-1

try this:

尝试这个：

char *process_name = "aaa\0";
memcpy((void *)argv[0], process_name, sizeof(process_name));

/* explain： The space allocated for argv[0] could be smaller than the name that you want to give and then you will be overwritting some other unrelated memory. argv[0] size could be just 2 and if your process name is "averylongprocessname" you will be overflowing argv[0]. You need to strlen(argv[0]) and use that in memcpy. thx @ecerulm

/ * explain：为argv [0]分配的空间可能小于你想要给出的名称，然后你将覆盖其他一些不相关的内存。 argv [0]大小可能只有2，如果你的进程名称是“averylongprocessname”，你将溢出argv [0]。你需要strlen（argv [0]）并在memcpy中使用它。 thx @ecerulm

* /

#1