# execve is not atomic, what is the exit state of the process when execve fails after throwing away the original process image?

- From: Steven Stewart-Gallus \<hidden\>
- Date: 2014-05-02 02:19:52 
- List: linux-api
- Page: https://ratatoskr.run/linux-api/2014/05/13328936
- Thread: https://ratatoskr.run/linux-api/2014/05/13328936/t.md

```text
Hello, can any kernel gurus please help me out with what happens on this weird
corner-case?


Quoting Rich Felker at http://ewontfix.com/14/:

> For common reasons it might fail, the execve syscall returns failure
> in the original process image, allowing the program to handle the
> error. However, failure of execve is not entirely atomic:
> 
>     The kernel may fail setting up the VM for the new process image
>     after the original VM has already been destroyed; the main
>     situation under which this would happen is resource exhaustion.
> 
>     Even after the kernel successfully sets up the new VM and
>     transfers execution to the new process image, it's possible to
>     have failures prior to the transfer of control to the actual
>     application program. This could happen in the dynamic linker
>     (resource exhaustion or other transient failures mapping required
>     libraries or loading configuration files) or libc startup
>     code. Using musl libc with static linking or even dynamic linking
>     with no additional libraries eliminates these failure cases, but
>     systemd is intended to be used with glibc.

execve is not atomic, what is the exit state of the process when
execve fails after throwing away the original process image?

I already have a hacky solution (see
https://gitorious.org/linted/linted/source/d25eac7a6fb58946ec8112771c0c56eb39fd1055:src/spawn/spawn.c)
to finding when a process fails before or at execve and now I want to
know what exit status is returned so I can give a nice error message.
```