Arguments, return values, and scope

Introduction to Bash Scripting

Alex Scriven

Data Scientist

Passing arguments into Bash functions

Passing arguments into functions is similar to how you pass arguments into a script. Using the $1 notation.

You also have access to the special ARGV properties we previously covered:

Each argument can be accessed via the $1, $2 notation.
$@ and $* give all the arguments in ARGV
$# gives the length (number) of arguments

Passing arguments example

Let's pass some file names as arguments into a function to demonstrate. We will loop through them and print them out.

function print_filename {
    echo "The first file was $1"

    for file in $@
    do
        echo "This file has name $file"
    done
}

print_filename "LOTR.txt" "mod.txt" "A.py"

The first file was LOTR.txt
This file has name LOTR.txt
This file has name mod.txt
This file has name A.py

Scope in programming

'Scope' in programming refers to how accessible a variable is.

'Global' means something is accessible anywhere in the program, including inside FOR loops, IF statements, functions etc.
'Local' means something is only accessible in a certain part of the program.

Why does this matter? If you try and access something that only has local scope - your program may fail with an error!

Scope in Bash functions

Unlike most programming languages (eg. Python and R), all variables in Bash are global by default.

function print_filename {
    first_filename=$1
}

print_filename "LOTR.txt" "model.txt"
echo $first_filename

LOTR.txt

Beware global scope may be dangerous as there is more risk of something unintended happening.

Restricting scope in Bash functions

You can use the local keyword to restrict variable scope.

function print_filename {
    local first_filename=$1
}

print_filename "LOTR.txt" "model.txt"
echo $first_filename

Q: Why wasn't there an error, just a blank line?

Answer: first_filename got assigned to the global first ARGV element ($1).

I ran the script with no arguments (bash script.sh) so this defaults to a blank element. So be careful!

Return values

We know how to get arguments in - how about getting them out?

The return option in Bash is only meant to determine if the function was a success (0) or failure (other values 1-255). It is captured in the global variable $?

Our options are:

Assign to a global variable
echo what we want back (last line in function) and capture using shell-within-a-shell

A return error

Let's see a return error:

function function_2 {
    echlo # An error of 'echo'
}

function_2 # Call the function
echo $? # Print the return value

script.sh: line 2: echlo: command not found
127

What happened?

There was an error when we called the function
- The script tried to find 'echlo' as a program but it didn't exist
The return value in $? was 127 (error)

Returning correctly

Let's correctly return a value to be used elsewhere in our script using echo and shell-within-a-shell capture:

function convert_temp {
    echo $(echo "scale=2; ($1 - 32) * 5 / 9" | bc)
}

converted=$(convert_temp 30)
echo "30F in Celsius is $converted C"

30F in Celsius is -1.11 C

See how we no longer create the intermediary variable?

Let's practice!

Introduction to Bash Scripting