Hands-on variable assignment

Open a new blank file and add the following to it:

#!/bin/bash

PI=3.14
VAR_A=10
VAR_B=$VAR_A
VAR_C=${VAR_B}

echo "Let's print 3 variables:"
echo $VAR_A
echo $VAR_B
echo $VAR_C

echo "We know this will break:"
echo "0. The value of PI is $PIabc"     # since PIabc is not declared, it will be empty string

echo "And these will work:"
echo "1. The value of PI is $PI"
echo "2. The value of PI is ${PI}"
echo "3. The value of PI is" $PI

echo "And we can make a new string"
STR_A="Bob"
STR_B="Jane"
echo "${STR_A} + ${STR_B} equals Bob + Jane"
STR_C=${STR_A}" + "${STR_B}
echo "${STR_C} is the same as Bob + Jane too!"
echo "${STR_C} + ${PI}"

exit 0

Press Save and exit to a terminal (open one if one isn't already open). Execute your script after applying the appropriate permissions, and you should see the following output:

Lets print 3 variables:
10
10
10
We know this will break:
0. The value of PI is 
And these will work:
1. The value of PI is 3.14
2. The value of PI is 3.14
3. The value of PI is 3.14
And we can make a new string
Bob + Jane equals Bob + Jane
Bob + Jane is the same as Bob + Jane too!
Bob + Jane + 3.14

First, we saw how we can use three variables, assign values to each of then, and print them. Secondly, we saw through a demonstration that the interpreter can break when concatenating strings (let's keep this in mind). Thirdly, we printed out our PI variable and concatenated it to a string using echo. Finally, we performed a few more types of concatenation, including a final version, which converts a numeric value and appends it to a string.