When I type ls -l $(echo file) output from bracket (which is just simple echo'ing) is taken and passed to external ls -l command. It equals to simple ls -l file.
When I type ls -l (echo file) we have error because one cannot nest () inside external command.
Can someone help me understand the difference between $() and () ?
$(cmd) substitutes the result of cmd as a string, whereas (cmd; cmd) run a list of commands in a subprocess.
If you want to put the output of one or more commands into a variable use the $( cmd ) form.
However if you want to run a number of commands and treat them as a single unit use the () form.
The latter is useful when you want to run a set of commands in the background:
(git pull; make clean; make all) &
variable=$(ls; ps) </br> echo $variable and eveyrting is correct , but </br> variable=(ls; ps) gives syntax error </br> (ls; ps) executes both commands </br> variable=(ls ps); echo $variable gives "ls" literallyThey are different, but there is a mnemonic relationship between them.
(...) is a command that starts a new subshell in which shell commands are run.
$(...) is an expression that starts a new subshell, whose expansion is the standard output produced by the commands it runs.
This is similar to another command/expression pair in bash: ((...)) is an arithmetic statement, while $((...)) is an arithmetic expression.
Those are different things.
$() evaluates an expression (executing a command) like `` (backticks)
> (echo ls)
ls
> $(echo ls)
file1 file2
> `echo ls`
file1 file2
> echo $(echo ls)
ls
echo $(echo ls)() is also used to create array in bash. See man bash:
Arrays are assigned to using compound assignments of the form name=(value1 ... valuen), where each value is of the form [subscript]=string.
$ arr=(a b c 'd f' e)
$ declare -p arr
declare -a arr=([0]="a" [1]="b" [2]="c" [3]="d f" [4]="e")
ls -l (echo file)to do?