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我写了一个简单的脚本，它接受任意数量的参数来演示 $@ 和 $* 之间的区别:
#!/bin/bash
echo "double quoted $* $@"
echo 'single quoted $* $@'
我在 CLI 上做的
$./stuff.sh a b c d e f dfs
这是打印出来的
double quoted a b c d e f dfs a b c d e f dfs
single quoted $* $@
既然它们相同，是否意味着 $@ 等于 $*？还是我遗漏了一点？
最佳答案
来自 Special Parameters in Bash Reference Manual
*
Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the IFS special variable. That is, "$*" is equivalent to "$1c$2c…", where c is the first character of the value of the IFS variable. If IFS is unset, the parameters are separated by spaces. If IFS is null, the parameters are joined without intervening separators.
@
Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word. That is, "$@" is equivalent to "$1" "$2" …. If the double-quoted expansion occurs within a word, the expansion of the first parameter is joined with the beginning part of the original word, and the expansion of the last parameter is joined with the last part of the original word. When there are no positional parameters, "$@" and $@ expand to nothing (i.e., they are removed).
举个例子更好:
$ d=(a b c)
$ for i in "${d[*]}"; do echo $i; done <---- it is a field all together
a b c
$ for i in "${d[@]}"; do echo $i; done <---- each item is a different field
a
b
c
关于bash - $@ 和 $* 之间的区别，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/19053064/