标题: bash shell : 怎么才能用程序比较出变量$@和$*的差别呀。我比较不出来。 [打印本页] 作者: wwgghh1976 时间: 2009-09-03 20:28 标题: bash shell : 怎么才能用程序比较出变量$@和$*的差别呀。我比较不出来。 真的弄不明白这两个变量有什么不同，希望高人能给点比较详细的解释。谢谢。作者: Henk2009 时间: 2009-09-06 11:40
$*
All of the positional parameters, seen as a single word
"$*" must be quoted.
$@
Same as $*, but each parameter is a quoted string, that is, the parameters are passed on intact, without
interpretation or expansion. This means, among other things, that each parameter in the argument list
is seen as a separate word.
-------
$* ===> "$1 $2...."
$@ ===> "$1" "$2" ....作者: wwgghh1976 时间: 2009-09-06 14:54 标题: 回复 #2 Henk2009 的帖子 我的e1.sh如下
#!/bin/sh
echo $*
或者
#!/bin/sh
echo $@
或者
#!/bin/sh
echo "$*"
或者
#!/bin/sh
echo "$@"
运行脚本：./e1.sh a b c "d e" 后的结果都是下面这样子
a b c d e作者: lijiecong 时间: 2009-09-06 21:19
=======test1.sh=====
#!/bin/sh
./test2.sh "$*"
./test2.sh "$@"
