Bash script execution with and without shebang in Linux and BSD

Question

How and who determines what executes when a Bash-like script is executed as a binary without a shebang?

I guess that running a normal script with shebang is

Answer 1

Since this happens in dash and dash is simpler, I looked there first.

Seems like exec.c is the place to look, and the relevant functionis are tryexec, which is called from shellexec which is called whenever the shell things a command needs to be executed. And (a simplified version of) the tryexec function is as follows:

STATIC void
tryexec(char *cmd, char **argv, char **envp)
{
        char *const path_bshell = _PATH_BSHELL;

repeat:

        execve(cmd, argv, envp);

        if (cmd != path_bshell && errno == ENOEXEC) {
                *argv-- = cmd;
                *argv = cmd = path_bshell;
                goto repeat;
        }
}

So, it simply always replaces the command to execute with the path to itself (_PATH_BSHELL defaults to "/bin/sh") if ENOEXEC occurs. There's really no magic here.

I find that FreeBSD exhibits identical behavior in bash and in its own sh.

The way bash handles this is similar but much more complicated. If you want to look in to it further I recommend reading bash's execute_command.c and looking specifically at execute_shell_script and then shell_execve. The comments are quite descriptive.

Answer 2

(Looks like Sorpigal has covered it but I've already typed this up and it may be of interest.)

According to Section 3.16 of the Unix FAQ, the shell first looks at the magic number (first two bytes of the file). Some numbers indicate a binary executable; #! indicates that the rest of the line should be interpreted as a shebang. Otherwise, the shell tries to run it as a shell script.

Additionally, it seems that csh looks at the first byte, and if it's #, it'll try to run it as a csh script.