CODEWITHSUNDEEP
bashparameters
Peter
Peter

Reputation: 333

Possible to pass literal character * as a command line argument in Bash?

Take the bash script wtf.sh

#!/bin/bash
echo $1
echo $2
echo $3

./wtf.sh 1 2 3 
1 
2
3

./wtf.sh * * * 



end

./wtf.sh 1 * * 
1
wtf.sh
wtf.sh 

./wtf.sh 1 "*" "*"
1
wtf.sh
wtf.sh

I'm aware that * is special parameter in Bash but does that mean it's impossible to pass the * by itself as a literal character from the command line? What I was expecting was 

./wtf.sh * * * 
* 
* 
*

How do I accomplish this?

Upvotes: 4

Views: 3833

Answers (2)

rahul.deshmukhpatil
rahul.deshmukhpatil

Reputation: 997

If you are passing * as a value of an command line option and extracting the command line option using getopt then you can use -S*, where -S is command line option and * is its value. Please note that there is no space between S and *. If you give space then path expansion happens automatically.

Upvotes: 0

rici
rici

Reputation: 241691

You can pass a literal * by:

  • Putting it in quotes: "*"

  • Putting it in apostrophes: '*'

  • Backslash escaping it: \*

Those are all equivalent (in this case).

The reason that didn't work is that your script has a problem: when you don't quote $1 (and $2 and $3), you are telling the shell to perform pathname expansion and word splitting on their contents. That's actually almost never what you want -- and it clearly isn't what you want in this case. So your script should read:

#!/bin/bash
echo "$1"
echo "$2"
echo "$3"

Here, the arguments must be surround with double quotes ("$1"), which allows parameter expansion to take place, but inhibits pathname expansion and word splitting.

Upvotes: 7

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