---
tags:
  - bash
  - shell
  - scripting
  - expansion
  - substitution
  - text
  - variable
  - output
  - execute
  - stdout
  - save
  - result
  - return
  - value
---

# Command substitution

    $( <COMMANDS> )

    ` <COMMANDS> `

The command substitution expands to the output of commands. These
commands are executed in a subshell, and their `stdout` data is what the
substitution syntax expands to.

All **trailing** newlines are removed (below is an example for a
workaround).

In later steps, **if not quoted**, the results undergo [word
splitting](../../syntax/expansion/wordsplit.md) and [pathname
expansion](../../syntax/expansion/globs.md). You have to remember that, because
the word splitting will also remove embedded newlines and other `IFS`
characters and break the results up into several words. Also you\'ll
probably get unexpected pathname matches. **If you need the literal
results, quote the command substitution!**

The second form `` `COMMAND` `` is more or less obsolete for Bash, since
it has some trouble with nesting (\"inner\" backticks need to be
escaped) and escaping characters. Use `$(COMMAND)`, it's also POSIX!

When you [call an explicit subshell](../../syntax/ccmd/grouping_subshell.md)
`(COMMAND)` inside the command substitution `$()`, then take care, this
way is **wrong**:

    $((COMMAND))

Why? because it collides with the syntax for [arithmetic
expansion](../../syntax/expansion/arith.md). You need to separate the command
substitution from the inner `(COMMAND)`:

    $( (COMMAND) )

## Specialities

When the inner command is only an input redirection, and nothing else,
for example

    $( <FILE )
    # or
    ` <FILE `

then Bash attempts to read the given file and act just if the given
command was `cat FILE`.

## A closer look at the two forms

In general you really should only use the form `$()`, it's
escaping-neutral, it's nestable, it's also POSIX. But take a look at
the following code snips to decide yourself which form you need under
specific circumstances:

**[Nesting]{.underline}**

Backtick form `` `...` `` is not directly nestable. You will have to
escape the \"inner\" backticks. Also, the deeper you go, the more escape
characters you need. Ugly.

    echo `echo `ls``      # INCORRECT
    echo `echo \`ls\``    # CORRECT
    echo $(echo $(ls))    # CORRECT

**[Parsing]{.underline}**

All is based on the fact that the backquote-form is simple character
substitution, while every `$()`-construct opens an own, subsequent
parsing step. Everything inside `$()` is interpreted as if written
normal on a commandline. No special escaping of **nothing** is needed:

    echo "$(echo "$(ls)")" # nested double-quotes - no problem

**[Constructs you should avoid]{.underline}**

It's not all shiny with `$()`, at least for my current Bash
(`3.1.17(1)-release`. :!: [**Update:** Fixed since `3.2-beta` together
with a misinterpretion of \'))\' being recognized as arithmetic
expansion \[by redduck666\]]{.underline}). This command seems to
incorrectly close the substitution step and echo prints \"ls\" and
\")\":

    echo $(
    # some comment ending with a )
    ls
    )

It seems that every closing \")\" confuses this construct. Also a (very
uncommon ;-)) construct like:

    echo $(read VAR; case "$var" in foo) blah ;; esac) # spits out some error, when it sees the ";;"

    # fixes it:
    echo $(read VAR; case "$var" in (foo) blah ;; esac) # will work, but just let it be, please ;-)

**[Conclusion:]{.underline}**

In general, the `$()` should be the preferred method:

-   it's clean syntax
-   it's intuitive syntax
-   it's more readable
-   it's nestable
-   its inner parsing is separate

## Examples

**To get the date:**

    DATE="$(date)"

**To copy a file and get `cp` error output:**

    COPY_OUTPUT="$(cp file.txt /some/where 2>&1)"

Attention: Here, you need to redirect `cp` `STDERR` to its `STDOUT`
target, because command substitution only catches `STDOUT`!

**Catch stdout and preserve trailing newlines:**

    var=$(echo -n $'\n'); echo -n "$var"; # $var == ""
    var=$(echo -n $'\n'; echo -n x); var="${var%x}"; echo -n "$var" # $var == "\n"

This adds \"x\" to the output, which prevents the trailing newlines of
the previous commands\' output from being deleted by \$().

By removing this \"x\" later on, we are left with the previous
commands\' output with its trailing newlines.

## See also

-   Internal: [Introduction to expansion and
    substitution](../../syntax/expansion/intro.md)
-   Internal: [Obsolete and deprecated syntax](../../scripting/obsolete.md)