Propagate all arguments in a bash shell script

Question

I am writing a very simple script that calls another script, and I need to propagate the parameters from my current script to the script I am executing.

For instance, my script name is foo.sh and calls bar.sh

foo.sh:

bar $1 $2 $3 $4

How can I do this without explicitly specifying each parameter?

Answer 1

Use "$@" instead of plain $@ if you actually wish your parameters to be passed the same.

Observe:

$ cat no_quotes.sh
#!/bin/bash
echo_args.sh $@

$ cat quotes.sh
#!/bin/bash
echo_args.sh "$@"

$ cat echo_args.sh
#!/bin/bash
echo Received: $1
echo Received: $2
echo Received: $3
echo Received: $4

$ ./no_quotes.sh first second
Received: first
Received: second
Received:
Received:

$ ./no_quotes.sh "one quoted arg"
Received: one
Received: quoted
Received: arg
Received:

$ ./quotes.sh first second
Received: first
Received: second
Received:
Received:

$ ./quotes.sh "one quoted arg"
Received: one quoted arg
Received:
Received:
Received: