112

I've got the following bash two scripts

a.sh:

#!/bin/bash
./b.sh 'My Argument'

b.sh:

#!/bin/bash
someApp $*

The someApp binary receives $* as 2 arguments ('My' and 'Argument') instead of 1.

I've tested several things:

  • Running someApp only thru b.sh works as expected
  • Iterate+echo the arguments in b.sh works as expected
  • Using $@ instead of $* doesn't make a difference
Improve this question
1
  • 7
    try someApp "$*" or someApp "$@" Commented Jun 13, 2013 at 18:13

1 Answer 1

Reset to default
168

$*, unquoted, expands to two words. You need to quote it so that someApp receives a single argument.

someApp "$*"

It's possible that you want to use $@ instead, so that someApp would receive two arguments if you were to call b.sh as

b.sh 'My first' 'My second'

With someApp "$*", someApp would receive a single argument My first My second. With someApp "$@", someApp would receive two arguments, My first and My second.

Improve this answer
Sign up to request clarification or add additional context in comments.

4 Comments

The key that is easy to miss is that "$@" needs to be quoted, it seems. $@ is not enough.
@miracle2k Correct. Unquoted, $@ and $* work identically.
@Matt That is something entirely different. "$@" is special in that it quotes each argument separately. It can result in more than one argument thus. "$something" is just quotes around whatever $something expands to and will always be a single argument.
The answer only works for me if I set IFS=$'\n'. No idea why.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.