What is the difference between ${var}, "$var", and "${var}" in the Bash shell?

Braces ($var vs. ${var})

In most cases, $var and ${var} are the same:

var=fooecho $var# fooecho ${var}# foo

The braces are only needed to resolve ambiguity in expressions:

var=fooecho $varbar# Prints nothing because there is no variable 'varbar'echo ${var}bar# foobar

Quotes ($var vs. "$var" vs. "${var}")

When you add double quotes around a variable, you tell the shell to treat it as a single word, even if it contains whitespaces:

var="foo bar"for i in "$var"; do # Expands to 'for i in "foo bar"; do...'    echo $i         #   so only runs the loop oncedone# foo bar

Contrast that behavior with the following:

var="foo bar"for i in $var; do # Expands to 'for i in foo bar; do...'    echo $i       #   so runs the loop twice, once for each argumentdone# foo# bar

As with $var vs. ${var}, the braces are only needed for disambiguation, for example:

var="foo bar"for i in "$varbar"; do # Expands to 'for i in ""; do...' since there is no    echo $i            #   variable named 'varbar', so loop runs once anddone                   #   prints nothing (actually "")var="foo bar"for i in "${var}bar"; do # Expands to 'for i in "foo barbar"; do...'    echo $i              #   so runs the loop oncedone# foo barbar

Note that "${var}bar" in the second example above could also be written "${var}"bar, in which case you don't need the braces anymore, i.e. "$var"bar. However, if you have a lot of quotes in your string these alternative forms can get hard to read (and therefore hard to maintain). This page provides a good introduction to quoting in Bash.

Arrays ($var vs. $var[@] vs. ${var[@]})

Now for your array. According to the bash manual:

Referencing an array variable without a subscript is equivalent to referencing the array with a subscript of 0.

In other words, if you don't supply an index with [], you get the first element of the array:

foo=(a b c)echo $foo# a

Which is exactly the same as

foo=(a b c)echo ${foo}# a

To get all the elements of an array, you need to use @ as the index, e.g. ${foo[@]}. The braces are required with arrays because without them, the shell would expand the $foo part first, giving the first element of the array followed by a literal [@]:

foo=(a b c)echo ${foo[@]}# a b cecho $foo[@]# a[@]

This page is a good introduction to arrays in Bash.

Quotes revisited (${foo[@]} vs. "${foo[@]}")

You didn't ask about this but it's a subtle difference that's good to know about. If the elements in your array could contain whitespace, you need to use double quotes so that each element is treated as a separate "word:"

foo=("the first" "the second")for i in "${foo[@]}"; do # Expands to 'for i in "the first" "the second"; do...'    echo $i              #   so the loop runs twicedone# the first# the second

Contrast this with the behavior without double quotes:

foo=("the first" "the second")for i in ${foo[@]}; do # Expands to 'for i in the first the second; do...'    echo $i            #   so the loop runs four times!done# the# first# the# second

TL;DR

All the examples you give are variations on Bash Shell Expansions. Expansions happen in a particular order, and some have specific use cases.

Braces as Token Delimiters

The ${var} syntax is primarily used for delimiting ambiguous tokens. For example, consider the following:

$ var1=foo; var2=bar; var12=12$ echo $var1212$ echo ${var1}2foo2

Braces in Array Expansions

The braces are required to access the elements of an array and for other special expansions. For example:

$ foo=(1 2 3)# Returns first element only.$ echo $foo1# Returns all array elements.$ echo ${foo[*]}1 2 3# Returns number of elements in array.$ echo ${#foo[*]}3

Tokenization

Most of the rest of your questions have to do with quoting, and how the shell tokenizes input. Consider the difference in how the shell performs word splitting in the following examples:

$ var1=foo; var2=bar; count_params () { echo $#; }# Variables are interpolated into a single string.$ count_params "$var1 $var2"1# Each variable is quoted separately, created two arguments.$ count_params "$var1" "$var2"2

The @ symbol interacts with quoting differently than *. Specifically:

1. $@ "[e]xpands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word."
2. In an array, "[i]f the word is double-quoted, ${name[*]} expands to a single word with the value of each array member separated by the first character of the IFS variable, and ${name[@]} expands each element of name to a separate word."

You can see this in action as follows:

$ count_params () { echo $#; }$ set -- foo bar baz $ count_params "$@"3$ count_params "$*"1

The use of a quoted expansion matters a great deal when variables refer to values with spaces or special characters that might prevent the shell from word-splitting the way you intend. See Quoting for more on how quoting works in Bash.

You need to distinguish between arrays and simple variables — and your example is using an array.

For plain variables:

• $var and ${var} are exactly equivalent.
• "$var" and "${var}" are exactly equivalent.

However, the two pairs are not 100% identical in all cases. Consider the output below:

$ var="  abc  def  "$ printf "X%sX\n" $varXabcXXdefX$ printf "X%sX\n" "${var}"X  abc  def  X$

Without the double quotes around the variable, the internal spacing is lost and the expansion is treated as two arguments to the printf command. With the double quotes around the variable, the internal spacing is preserved and the expansion is treated as one argument to the printf command.

With arrays, the rules are both similar and different.

• If groups is an array, referencing $groups or ${groups} is tantamount to referencing ${groups[0]}, the zeroth element of the array.
• Referencing "${groups[@]}" is analogous to referencing "$@"; it preserves the spacing in the individual elements of the array, and returns a list of values, one value per element of the array.
• Referencing ${groups[@]} without the double quotes does not preserve spacing and can introduce more values than there are elements in the array if some of the elements contain spaces.

For example:

$ groups=("abc def" "  pqr  xyz  ")$ printf "X%sX\n" ${groups[@]}XabcXXdefXXpqrXXxyzX$ printf "X%sX\n" "${groups[@]}"Xabc defXX  pqr  xyz  X$ printf "X%sX\n" $groupsXabcXXdefX$ printf "X%sX\n" "$groups"Xabc defX$

Using * instead of @ leads to subtly different results.

See also How to iterate over the arguments in a bash script.