假设您想要找到与gcc安装位置对应的lib目录。你有一个选择:

libdir=$(dirname $(dirname $(which gcc)))/lib

libdir=`dirname \`dirname \\\`which gcc\\\`\``/lib

第一种方法比第二种简单——使用第一种方法。

最主要的一个是嵌套它们的能力，命令中的命令，而不会失去理智，试图弄清楚某种形式的转义是否对反节拍有效。

一个例子，虽然有点做作:

deps=$(find /dir -name $(ls -1tr 201112[0-9][0-9]*.txt | tail -1l) -print)

它会给你一个/dir目录树中所有与2011年12月以来最早的文本文件同名的文件的列表(a)。

另一个例子是获取父目录的名称(而不是完整路径):

pax> cd /home/pax/xyzzy/plugh
pax> parent=$(basename $(dirname $PWD))
pax> echo $parent
xyzzy

(a)既然那个特定的命令可能实际上不起作用，我还没有测试它的功能。所以，如果你投我的反对票，你已经失去了意图:-)这只是一个说明你如何嵌套，而不是作为一个没有bug的生产准备片段。

来自man bash:

       $(command)
or
       `command`

Bash performs the expansion by executing command and replacing the com-
mand  substitution  with  the  standard output of the command, with any
trailing newlines deleted.  Embedded newlines are not deleted, but they
may  be  removed during word splitting.  The command substitution $(cat
file) can be replaced by the equivalent but faster $(< file).

When the old-style backquote form of substitution  is  used,  backslash
retains  its  literal  meaning except when followed by $, `, or \.  The
first backquote not preceded by a backslash terminates the command sub-
stitution.   When using the $(command) form, all characters between the
parentheses make up the command; none are treated specially.

反引号('…')是只有最古老的不兼容POSIX的boure -shell才需要的遗留语法，而$(…)是POSIX，出于以下几个原因更受欢迎:

Backslashes (\) inside backticks are handled in a non-obvious manner: $ echo "`echo \\a`" "$(echo \\a)" a \a $ echo "`echo \\\\a`" "$(echo \\\\a)" \a \\a # Note that this is true for *single quotes* too! $ foo=`echo '\\'`; bar=$(echo '\\'); echo "foo is $foo, bar is $bar" foo is \, bar is \\ Nested quoting inside $() is far more convenient: echo "x is $(sed ... <<<"$y")" instead of: echo "x is `sed ... <<<\"$y\"`" or writing something like: IPs_inna_string=`awk "/\`cat /etc/myname\`/"'{print $1}' /etc/hosts` because $() uses an entirely new context for quoting which is not portable as Bourne and Korn shells would require these backslashes, while Bash and dash don't. Syntax for nesting command substitutions is easier: x=$(grep "$(dirname "$path")" file) than: x=`grep "\`dirname \"$path\"\`" file` because $() enforces an entirely new context for quoting, so each command substitution is protected and can be treated on its own without special concern over quoting and escaping. When using backticks, it gets uglier and uglier after two and above levels. Few more examples: echo `echo `ls`` # INCORRECT echo `echo \`ls\`` # CORRECT echo $(echo $(ls)) # CORRECT It solves a problem of inconsistent behavior when using backquotes: echo '\$x' outputs \$x echo `echo '\$x'` outputs $x echo $(echo '\$x') outputs \$x Backticks syntax has historical restrictions on the contents of the embedded command and cannot handle some valid scripts that include backquotes, while the newer $() form can process any kind of valid embedded script. For example, these otherwise valid embedded scripts do not work in the left column, but do work on the rightIEEE: echo ` echo $( cat <<\eof cat <<\eof a here-doc with ` a here-doc with ) eof eof ` ) echo ` echo $( echo abc # a comment with ` echo abc # a comment with ) ` ) echo ` echo $( echo '`' echo ')' ` )

Therefore the syntax for $-prefixed command substitution should be the preferred method, because it is visually clear with clean syntax (improves human and machine readability), it is nestable and intuitive, its inner parsing is separate, and it is also more consistent (with all other expansions that are parsed from within double-quotes) where backticks are the only exception and ` character is easily camouflaged when adjacent to " making it even more difficult to read, especially with small or unusual fonts.

来源:为什么$(…)比'…”(引号)?在BashFAQ

参见:

POSIX标准节“2.6.3命令替换” 包含$()语法的POSIX原理 命令替换 Bash-hackers:命令替换

在2021年，有必要提到一个奇怪的事实，作为对其他答案的补充。

Microsoft DevOps YAML管道“脚本”可能包含Bash任务。但是，符号$()用于引用YAML上下文中定义的变量，因此在这种情况下，应该使用反引号来捕获命令的输出。

在将脚本代码复制到YAML脚本时，这主要是一个问题，因为DevOps预处理器对不存在的变量非常宽容，所以不会出现任何错误消息。

POSIX标准定义了命令替换的$(命令)形式。目前使用的大多数shell都是POSIX兼容的，并且支持这种首选形式，而不是古老的反刻度符号。Shell Language文档的命令替换部分(2.6.3)描述了以下内容:

Command substitution allows the output of a command to be substituted in place of the command name itself.  Command substitution shall occur when the command is enclosed as follows: $(command) or (backquoted version): `command` The shell shall expand the command substitution by executing command in a subshell environment (see Shell Execution Environment) and replacing the command substitution (the text of command plus the enclosing "$()" or backquotes) with the standard output of the command, removing sequences of one or more <newline> characters at the end of the substitution. Embedded <newline> characters before the end of the output shall not be removed; however, they may be treated as field delimiters and eliminated during field splitting, depending on the value of IFS and quoting that is in effect. If the output contains any null bytes, the behavior is unspecified. Within the backquoted style of command substitution, <backslash> shall retain its literal meaning, except when followed by: '$' , '`', or <backslash>. The search for the matching backquote shall be satisfied by the first unquoted non-escaped backquote; during this search, if a non-escaped backquote is encountered within a shell comment, a here-document, an embedded command substitution of the $(command) form, or a quoted string, undefined results occur. A single-quoted or double-quoted string that begins, but does not end, within the "`...`" sequence produces undefined results. With the $(command) form, all characters following the open parenthesis to the matching closing parenthesis constitute the command. Any valid shell script can be used for command, except a script consisting solely of redirections which produces unspecified results. The results of command substitution shall not be processed for further tilde expansion, parameter expansion, command substitution, or arithmetic expansion. If a command substitution occurs inside double-quotes, field splitting and pathname expansion shall not be performed on the results of the substitution. Command substitution can be nested. To specify nesting within the backquoted version, the application shall precede the inner backquotes with <backslash> characters; for example: \`command\` The syntax of the shell command language has an ambiguity for expansions beginning with "$((", which can introduce an arithmetic expansion or a command substitution that starts with a subshell. Arithmetic expansion has precedence; that is, the shell shall first determine whether it can parse the expansion as an arithmetic expansion and shall only parse the expansion as a command substitution if it determines that it cannot parse the expansion as an arithmetic expansion. The shell need not evaluate nested expansions when performing this determination. If it encounters the end of input without already having determined that it cannot parse the expansion as an arithmetic expansion, the shell shall treat the expansion as an incomplete arithmetic expansion and report a syntax error. A conforming application shall ensure that it separates the "$(" and '(' into two tokens (that is, separate them with white space) in a command substitution that starts with a subshell. For example, a command substitution containing a single subshell could be written as: $( (command) )