CODEWITHSUNDEEP
linuxbash
hel
hel

Reputation: 611

When does a variable add $ in bash

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

Upvotes: 0

Views: 75

Answers (4)

chepner
chepner

Reputation: 532313

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.

Upvotes: 2

sjsam
sjsam

Reputation: 21965

I thought @slaks' [ answer ] wouldn't be complete without this :

When not to add $ for a variable

  1. With The Double-Parentheses [ Construct ]

    x=5;
(( x++ )) # fine, note this construct accept $x form too.

  2. When using export

    var=stuff
export var #fine

  3. 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.

Upvotes: 3

SLaks
SLaks

Reputation: 888185

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.

Upvotes: 4

Syed
Syed

Reputation: 61

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.

Upvotes: 0

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