Pass bash script parameters to sub-process unchanged

Question

I want to write a simple bash script that will act as a wrapper for an executable. How do I pass all the parameters that script receives to the executable? I tried

/the/exe $@

but this doesn't work with quoted parameters, eg.

./myscript "one big parameter"

runs

/the/exe one big parameter

which is not the same thing.

Answer 1

You have to put the $@ in quotes:

/the/exe "$@"

Answer 2

When a shell script wraps around an executable, and if you do not want to do anything after the executable completes (that's a common case for wrapper scripts, in my experience), the correct way to call the executable is:

exec /the/exe "$@"

The exec built-in tells the shell to just give control to the executable without forking.

Practically, that prevents a useless shell process from hanging around in the system until the wrapped process terminates.

That also means that no command can be executed after the exec command.