$@和$*的问题!!!

晨想

这样看起来似乎更明了。

$ bash -x i
+ set -- 'First Arg' Second Third:Arg '' 'Fifth: :Arg'
+ IFS=:
+ var1='First Arg Second Third Arg  Fifth   Arg'
+ var2='First Arg:Second:Third:Arg::Fifth: :Arg'
+ echo ----------
----------
+ echo 'First Arg Second Third Arg  Fifth   Arg'
First Arg Second Third Arg  Fifth   Arg
+ echo 'First Arg:Second:Third:Arg::Fifth: :Arg'
First Arg:Second:Third:Arg::Fifth: :Arg
+ echo ----------
----------
+ IFS=
+ var1='First Arg Second Third:Arg  Fifth: :Arg'
+ var2='First ArgSecondThird:ArgFifth: :Arg'
+ echo ----------
----------
+ echo 'First Arg Second Third:Arg  Fifth: :Arg'
First Arg Second Third:Arg  Fifth: :Arg
+ echo 'First ArgSecondThird:ArgFifth: :Arg'
First ArgSecondThird:ArgFifth: :Arg

yfreedom

看了3楼yongjian兄的解释,但还是不明白下面这段代码啊,怎么是这个结果那??

望各位为我解惑啊

IFS=

var1=$*

echo "$var1"              # First Arg Second Third:Arg   FFiifftthh::  ::AArrg

yongjian

Post by yfreedom
看了3楼yongjian兄的解释,但还是不明白下面这段代码啊,怎么是这个结果那??

望各位为我解惑啊

IFS=

var1=$*

echo "$var1"              # First Arg Second Third:Arg   FFiifftthh::  ::AArrg

the result is

1. 'First ArgSecondThird:ArgFifth: :Arg'

复制代码

yfreedom

这论坛真学习linux得好地方啊，我嗷嗷喜欢啊