UNIX中的$@和$*有什么区别？

UNIX中的$@和$*有什么区别？当在脚本中回显时,它们似乎产生相同的输出.

一个区别在于它们如何处理输出上的IFS变量.

#!/bin/sh
echo "unquoted asterisk " $*
echo "quoted asterisk $*"
echo "unquoted at " $@
echo "quoted at $@"
IFS="X"
echo "IFS is Now $IFS"
echo "unquoted asterisk " $*
echo "quoted asterisk $*"
echo "unquoted at " $@
echo "quoted at $@"

如果你这样运行：./demo abc def ghi,你得到这个输出：

unquoted asterisk abc def ghi
quoted asterisk abc def ghi
unquoted at abc def ghi
quoted at abc def ghi
IFS is Now X
unquoted asterisk abc def ghi
quoted asterisk abcXdefXghi
unquoted at abc def ghi
quoted at abc def ghi

请注意,(仅)“IFS”更改为“X”后,“引用的星号”行显示每个“单词”之间的X.如果IFS的值包含多个字符,则仅使用第一个字符用于此目的.

此功能也可用于其他阵列：

$array=(123 456 789)
$saveIFS=$IFS; IFS="|"
$echo "${array[*]}"
123|456|789
$IFS=$saveIFS

原文地址：https://www.jb51.cc/bash/386635.html