How to pass arguments to function in a bash script?

You are running the addition_example function with no arguments. Therefore, the $1 and $2 variables are empty and what you're actually executing is just result = $((+)). This gives precisely the error you mention:

$ result = $((+))
bash: +: syntax error: operand expected (error token is "+")

When you run source 20 20, the shell will source, passing it 20 and 20 as arguments. However, doesn't actually contain any commands, it only has a function declaration. So the arguments are ignored. Then, in a subsequent command, you run addition_example but with no arguments, so you get the error described above.

Also, you have a syntax error. You can't have spaces around the assignment operator (=) in shell scripts. You should change your script to:

function addition_example(){
    echo "Addition of the supplied arguments = $result"

And then run the function with the desired arguments:

$ source; addition_example 20 20
Addition of the supplied arguments = 40

You have a few minor issues with the code.

Try this:


function addition_example(){
    local num1=$1
    local num2=$2
    local result=$((num1+num2))
    echo "Addition of the supplied arguments = $result"
addition_example "$1" "$2"
  • The function was being declared in the script but when you simply execute the script it would not execute the function.
  • The arguments passed to the script (as positional parameters) will not be passed to the function unless specified.
  • You cannot have spaces around the = in variable assignment with bash
  • Your echo statement should be quoted, any variable should be quoted.

There are several issues with the script.

Use Shellcheck

Some could be detected by using The first problem it shows is:

In dd line 4:
result = $(($1+$2))
       ^-- SC1068: Don't put spaces around the = in assignments.

Quote variables

You have a variable expansion unquoted. You should quote all variable expansions unless you really want that glob & split operations are needed. Change the echo line to:

echo "Addition of the supplied arguments = $result"

Do not use both function and () at the same time

The definition of the function could be either of this lines:

function add { 
add () {

The second method is more portable (runs on more shells).

Call the function with arguments

There are two methods to give parameters to the function.

Source it (not recommended)

You could source the script inside your present interactive shell:

$ source ./
$ addition_example 20 20
Addition of the supplied arguments = 40

Do not use $20 as the $ is a reserved character for the shell. It will be parsed by the shell as the second parameter $2 (in the present shell) and append a 0 after it.

However, this is changing the running shell with what is inside the script. Not something that you usually want.

Also, sourcing an script makes no use of the shebang line (the first line of the script). That will allow for any running shell in the interactive shell (sh,ksh,bash,zsh) to actually execute the code. If the code is written to be interpreted and executed with a specific shell.

Execute it

All you need to actually execute the script (as it should be) is:

  • Make the script executable chmod u+x
  • Include a shebang line (you already have one)
  • Do call the function.

Edited script


    echo "Addition of the supplied arguments = $result"

addition_example "$@"

And execute it as:

$ ./ 20 20
Addition of the supplied arguments = 40

The "$@" get expanded to the list of arguments supplied to the script.

Improve robustness

To make your script more robust, test if the arguments to the function are empty (which will print an error message):