Bash Shebang for dummies?

Question

I have some bash scripts I have setup that mostly use

#!/bin/bash

but I regularly come across some that look like

#!/bin/bash -e
#!/bin/bash -x
#!/bin/bash -ex

and so on.

Can someone explain the meaning and benefits of these shebang options and whether they apply to other shebangs?

Answer 1

If a script /path/to/foo begins with #!/bin/bash, then executing /path/to/foo arg1 arg2 is equivalent to executing /bin/bash /path/too/foo arg1 arg2. If the shebang line is #!/bin/bash -ex, it is equivalent to executing /bin/bash -ex /path/too/foo arg1 arg2. This feature is managed by the kernel.

Note that you can portably have only one argument on the shebang line: some unices (such as Linux) only accept one argument, so that #!/bin/bash -e -x would lead to bash receiving the single five-character argument -e -x (a syntax error) rather than two arguments -e and -x.

For the Bourne shell sh and derived shells such as POSIX sh, bash, ksh, and zsh:

• -e means that if any command fails (which it indicates by returning a nonzero status), the script will terminate immediately.
• -x causes the shell to print an execution trace.

Other programs may understand these options but with different meanings.

Answer 2

They are options passed to bash see help set for more info, in this case:

-x  Print commands and their arguments as they are executed.
-e  Exit immediately if a command exits with a non-zero status.