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Passing all arguments in Bash

Author:JIYIK Last Updated:2025/03/21 Views:

In Bash scripts, we mainly use the syntax $1 $2 $3 ... $Nto pass variables in functions. Here N is a non-negative integer such as 1, 2, 3, ....

But there is a shortcut to accomplish the same task of passing all the arguments. This article will discuss how to pass all the arguments in a function.

Furthermore, we will see necessary examples and explanations to make the topic easier to understand.


Passing all arguments in Bash

To do this, we will use $@to pass all the parameters. Let's see the following sample code.

My_Friends () {
echo "Your friend's names are: "$@""
}

My_Friends Alen Walker John

In the above example, we have passed three arguments in the function named My_Friends. We use $@ to receive all the functions in the function.

Now, when you run the above example, you will get the output as shown below.

Your friend's names are: Alen Walker John

Now the traditional way to perform the same task is as follows.

My_Friends () {
echo "Your friend's names are: $1 $2 $3"
}

My_Friends Alen Walker John

In the above code, we have included all the parameters manually. This will also give the same output as the example we provided earlier.

请注意, if you have to handle all the parameters individually, you cannot include them manually. For example, if you want to perform an add operation with all the passed parameters like our example below, then you cannot use this shortcut.

Let's check the following example.

Add () {
echo The sum is: $(( $1+$2+$3 ))
}

Add 1 2 3

As we have discussed, you have to mention all the parameters individually in the recently shared example. The output of the above program will be as shown below.

The sum is: 6

All the codes used in this article are written in Bash. It will only work in Linux Shell environment.

For reprinting, please send an email to 1244347461@qq.com for approval. After obtaining the author's consent, kindly include the source as a link.

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