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。

wwgghh1976

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。

Henk2009

$*
All of the positional parameters, seen as a single word
"$*" 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" ....

wwgghh1976

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

lijiecong

=======test1.sh=====
#!/bin/sh
./test2.sh "$*"
./test2.sh "$@"


======test2.sh==========
#!/bin/sh

for((i=0; i<$#;i++))
do
    echo $1
   shift
done



运行   ./test1.sh 1 2 3 4 5就可以看出来了

wwgghh1976