Change process name without changing argv[0] in Linux

Question

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.

Adopted

possible duplicate of How to name a thread in Linux? – Invertase

"The documentation says PR_SET_NAME sets the process name; but that documentation is wrong - it does actually set the thread name. Now "top" and "ps -L" show the thread name." -- user9876 Apr 22 '09 at 17:39 – Invertase

PR_SET_NAME flag is supported since Linux 2.6.9 – Seitz

Answer 1

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

• modify argv[0] changes information in /proc/$pid/cmdline
• invoking prctl(PR_SET_NAME) changes information in /proc/$pid/status

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

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).

Hope the answer helps.

Truthful

Answer 2

try this:

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

*/

Demonography

I think that the \0 is unnecessary - isn't it already silently appended to a literal string of characters? – Brozak

@Chap: it's true for strcpy, not for memcpy. – Leadsman

Seems that you have to overwrite all characters of origin argv[0], is it true? So you can't have process names bigger than original name – Contributory

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. – Sallee

The \0 is unnecessary. Also, sizeof is incorrect here as it would return the size of a pointer (typically 4 or 8 bytes). You want something more like strncpy(argv[0], process_name, strlen(process_name) or memcpy(argv[0], process_name, (strlen(argv[0]) < strlen(process_name)+1) ? strlen(argv[0]) : strlen(process_name)+1) – Angiosperm