Bash 数组创建：("$@") vs ($@)
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Bash array creation: ("$@") vs ($@)
提问by Roxana
I am running a script with: ./some_script arg1 arg2 "multiple words arg3" arg4. I want to explode arguments ($@) into an array. This snippet works just for arguments without spaces:
我运行一个脚本：./some_script arg1 arg2 "multiple words arg3" arg4。我想将参数 ( $@) 分解成一个数组。此代码段仅适用于没有空格的参数：
arr=($@)
If I want to store the correct arguments into array I must use:
如果我想将正确的参数存储到数组中，我必须使用：
arr=("$@")
Why should I enclose $@in quotes?
为什么我应该$@用引号引起来？
I think this has something to do with parameter expansion and special parameters, but I don't think I got it well.
我认为这与参数扩展和特殊参数有关，但我认为我做得不好。
采纳答案by Gordon Davisson
In the shell, whenever a variable (including special parameters like $@) in referenced without double-quotes, the value goes through word splitting and wildcard expansion after it's expanded. For example:
在shell中，每当一个变量（包括特殊参数，如$@）不带双引号引用时，该值在扩展后会经历分词和通配符扩展。例如：
$ var="FOO * BAR"
$ printf "%s\n" "$var"
FOO * BAR
$ printf "%s\n" $var
FOO
Desktop
Documents
Downloads
Library
Movies
Music
Pictures
Public
BAR
In the second case, the variable value "FOO * BAR" got split into separate words ("FOO", "*", and "BAR"), and then the "*" was expanded into a list of matching files. This is why you almost always want to put variable references in double-quotes.
在第二种情况下，变量值“FOO * BAR”被拆分为单独的词（“FOO”、“*”和“BAR”），然后将“*”扩展为匹配文件的列表。这就是为什么您几乎总是希望将变量引用放在双引号中。
The same thing applies to $@-- if it's not in double-quotes, it's expanded into the list of parameters and then each one of them is subjected to that same word splitting and wildcard expansion that $varwent through above. If it's in double-quotes, the parameter values are left unmolested.
同样的事情适用于$@- 如果它不是双引号，它会扩展到参数列表中，然后每个参数都受到与上面相同的分词和通配符扩展$var。如果是双引号，则参数值不会受到干扰。
BTW, there is another way to get the parameters: $*. This differs from $@in that it sticks all of the parameter values together with spaces between them (while $@maintains each parameter as a separate word). In double-quotes, "$*"gives a single word consisting of all parameters. Without double-quotes, $*sticks all the parameters together, then re-splits them (maybe at the same places, maybe not), and does wildcard expansion. Probably not what you wanted.
顺便说一句，还有另一种获取参数的方法：$*. 这与$@它的不同之处在于它将所有参数值粘在一起并在它们之间使用空格（同时$@将每个参数保持为一个单独的词）。在双引号中，"$*"给出一个包含所有参数的单词。没有双引号，将$*所有参数粘在一起，然后重新拆分它们（可能在相同的地方，也可能不在），并进行通配符扩展。可能不是你想要的。
回答by Gordon Davisson
If you didn't surround it with quotes, "multiple words arg3"would be further expanded to multiplewordsarg3- ie. the quotes preserve special characters after the variable has been expanded.
如果您没有用引号"multiple words arg3"将其括起来，则将进一步扩展为multiplewordsarg3- 即。引号在变量扩展后保留特殊字符。
In other words, when you don't surround a variable expansion with quotes, what the variable expands to will further be expanded, which would in this case eliminate the double quotes around the third argument.
换句话说，当你没有用引号包围变量扩展时，变量扩展到的内容将被进一步扩展，在这种情况下，这将消除第三个参数周围的双引号。
