$ *和$ @之间的差 [英] The difference between $* and $@

问题描述

可能重复：结果
  什么区别＆ldquo的区别; $ @＆rdquo;的和＆ldquo; $ * rdquo;的在bash？

有关年，几十个场合，我已经使用 $之间犹豫* 和 $ @ 中shell脚本。看了一遍又一遍的Bash的联机帮助的适用部分，尝试过两种 $ * 和 $ @ ，我更以下完全无法理解两个变量之间的应用程序的实际差别。你可以告诉我，好吗？

For years and on dozens of occasions, I have hesitated between the use of $* and $@ in shell scripts. Having read the applicable section of Bash's manpage over and over again, having tried both $* and $@, I more or less completely fail to understand the practical difference of application between the two variables. Can you enlighten me, please?

我一直使用 $ * 最近，但不要问我为什么。我不知道为什么，因为我不知道为什么 $ @ 即使存在，除非作为 $ *几乎确切的代名词。

I have been using $* recently, but don't ask me why. I don't know why, because I don't know why $@ even exists, except as an almost exact synonym for $*.

有没有实际的区别吗？

（我个人倾向于使用猛砸，但仍不可知对于外壳的选择。我的问题是不特定为据我所知猛砸。）

(I personally tend to use Bash, but remain agnostic regarding the choice of shell. My question is not specific to Bash as far as I know.)

推荐答案

未加引号的，没有任何区别 - 他们扩展到所有的争论和他们相应的分割。所不同的引用来的时候。 $ @扩展到正确引用参数和$ *把所有的参数到一个单一的参数。借此例如：

Unquoted, there is no difference -- they're expanded to all the arguments and they're split accordingly. The difference comes when quoting. "$@" expands to properly quoted arguments and "$*" makes all arguments into a single argument. Take this for example:

#!/bin/bash

function print_args_at {
    printf "%s\n" "$@"
}

function print_args_star {
    printf "%s\n" "$*"
}

print_args_at "one" "two three" "four"
print_args_star "one" "two three" "four"

然后：

$ ./printf.sh 

one
two three
four

one two three four

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