When does a variable add $ in bash - linux

Question

I'm recently learning bash and confused when a variable would add $. I find code like:
i=1
while [ $i -le 10 ]
do
echo "$n * $i = `expr $i \* $n`"
i=`expr $i + 1`
done

Answer 1

The $ substitutes the variable. Writing $i will insert the value of i, no matter where you write it.
If you want to assign to the variable, that obviously makes no sense.

Answer 2

I thought #slaks' [ answer ] wouldn't be complete without this :
When not to add $ for a variable
With The Double-Parentheses [ Construct ]
x=5;
(( x++ )) # fine, note this construct accept $x form too.
When using export
var=stuff
export var #fine
When using declare
declare -a arry # fine
When not omit $
As #rici pointed out in the comment below:
you can leave out the $ in any arithmetic context, not just ((...))
and $((...)) ... For example, if arr is an array (not associative), then
${arr[x++]} is also fine.
Consider
# You wanted to create an associative array 'test' but you forgot to do
# declare -A test , Now below
test[foo]=bar # is foo a variable or a key, the reader isn't clear
# creates a simple array
echo ${test[foo]} # is foo a variable or a key?
bar
declare -p test
declare -a test='([0]="bar")'
# What happened?
# Since foo was not set at the point when 'test[foo]=bar' was called,
# bash substituted it with zero
# I meant to say test[foo]=bar hides an error.

Answer 3

A key thing to remember is that variables are never passed around in shell, only values. When you call something like
echo "$foo"
you might think that echo receives $foo, then looks at its value. Instead, the shell first expands $foo to the value of foo, then passes that value to echo.
The dollar sign is used to introduce any such parameter expansion, where the value of a parameter is needed. Consider:
$ foo=10
$ echo foo
foo
$ echo $foo
10
From the perspective of the echo command, there is no difference between echo $foo and echo 10; in both cases, the value passed to echo is 10.

Answer 4

To set a value to a variable in bash you can do
a=10
Where as to access the value of that variable you need to use
echo $a which mean $ is a symbol used to access the value of a variable.
i=1 ---> setting variable as 1
while [ $i -le 10 ] ---> simple while statement which loops till value of i is less that 10
do
echo "$n * $i = `expr $i \* $n`" ---> This is a syntax error bcoz value of n is never assigned
i=`expr $i + 1` ---> This line add's one to value of i
done ---> terminate while loop
Hope that explains you.