How does bash execute an ELF file?

Question

When I'm on my Linux Box I use bash as a shell. Now I wondered how bash handles the execution of an ELF file, that is when I type ./program and program is an ELF file. I grepped the bash-4.3.tar.gz, there does not seem to be some sort of magic number parser to find out if the file is an ELF nor did I find an exec() syscall.

How does the process work? How does bash pass the execution of the ELF to the OS?

Solution

Bash knows nothing about ELF. It simply sees that you asked it to run an external program, so it passes the name you gave it as-is to execve(2). Knowledge of things like executable file formats, shebang lines, and execute permissions lives behind that syscall, in the kernel.

(It is the same for other shells, though they may choose to use another function in the exec(3) family instead.)

In Bash 4.3, this happens on line 5195 of execute_cmd.c in the shell_execve() function.

If you want to understand Linux at the source code level, I recommend downloading a copy of Research Unix V6 or V7, and going through that rather than all the complexity that is in the modern Linux systems. The Lions Book is a good guide to the code.

V7 is where the Bourne shell made its debut. Its entire C source code is just a bit over half the size of just that one C file in Bash. The Thompson shell in V6 is nearly half the size of the original Bourne shell. Yet, both of these simpler shells do the same sort of thing as Bash, and for the same reason. (It appears to be an execv(2) call from texec() in the Thompson shell and an execve() call from execs() in the Bourne shell's service.c module.)