How to change a command line argument in Bash?

Question

Is there a way to change the command line arguments in a Bash script? For example, a Bash script is invoked like this:

./foo arg1 arg2  

Is there a way to change the value of arg1 within the script? Something like:

$1="chintz"

Answer 1

You have to reset all arguments. To change e.g. argument $3:

$ set -- "${@:1:2}" "new_arg3" "${@:4}"

Basically you set all arguments to their current values, except for the one(s) that you want to change. set -- is also specified by POSIX 7.

The "${@:1:2}" notation is expanded to the two (hence the 2 in the notation) positional arguments starting from offset 1 (i.e. $1). It is a shorthand for "$1" "$2" in this case, but it is much more useful when you want to replace e.g. "${17}".

Answer 2

Optimising for legibility and maintainability, you may be better off assigning $1 and $2 to more meaningful variables (I don't know, input_filename = $1 and output_filename = $2 or something) and then overwriting one of those variables (input_filename = 'chintz'), leaving the input to the script unchanged, in case it is needed elsewhere.

Answer 3

I know this is an old one but I found the answer by thkala very helpful, so I have used the idea and expanded on it slightly to enable me to add defaults for any argument which has not been defined - for example:

    # set defaults for the passed arguments (if any) if not defined.
    #
    arg1=${1:-"default-for-arg-1"}
    arg2=${2:-"default-for-arg-2"}
    set -- "${arg1}" "${arg2}"
    unset arg1 arg2

I hope this is of use to someone else.

Answer 4

How to change command-line arguments $1, $2, etc. in bash using set --

As an addition to the main answer on Ask Ubuntu by @Radu Rădeanu here, and to the main answer by @thkala on Stack Overflow here, I'd like to add a really simple example to set multiple arguments at once like this:

# set arguments `$1`, `$2`, and `$3` to these respective values
set -- "one" "two" "three"
echo "$1 $2 $3"

Output:

one two three

If you have additional arguments after the arguments you want to change, you can keep them by slicing them onto the end, like this:

set -- "one" "two" "three" "${@:4}"

I explain the "${@:4}" slicing syntax in my answer here: Unix & Linux: Bash: slice of positional parameters. See that answer for much more detail on bash array slicing. Here is a small snippet from it:

# array slicing basic format 1: grab a certain length starting at a certain
# index
echo "${@:2:5}"
#         │ │
#         │ └────> slice length
#         └──────> slice starting index 

# array slicing basic format 2: grab all remaining array elements starting at a
# certain index through to the end
echo "${@:2}"
#         │
#         │
#         └──────> slice starting index 

So, in the main answer by @thkala, which contains this:

set -- "${@:1:2}" "new_arg3" "${@:4}"

...the explanation is that "${@:1:2}" grabs the first 2 arguments, "new_arg3" inserts that as a new argument 3, and "${@:4}" grabs all arguments from 4 to the end, thereby changing only argument 3 and keeping all other arguments as they were.

For help on the set built-in bash command, see help set or look online here: https://www.gnu.org/savannah-checkouts/gnu/bash/manual/bash.html#The-Set-Builtin

help set shows the following for the -- argument:

 --  Assign any remaining arguments to the positional parameters.
     If there are no remaining arguments, the positional parameters
     are unset.

See also

1. My shorter answer on Ask Ubuntu here: How to change the value of an argument in a script?