shell $@ 变量赋值和迭代 [英] shell $@ variable assignment and iteration

问题描述

function f() {
    for arg; do
        echo "$arg"
    done
}

f 'hello world' 'second arg' '3rdarg'

这很好用:

hello world
second arg
3rdarg

但是我什么时候将 $@ 分配给某个变量，然后它出错了:

but when do I assign $@ to some variable, and then it goes wrong:

function f() {
    local args="$@"
    for arg in "$args"; do
        echo "$arg"
    done
}

输出:

hello world second arg 3rdarg

当我取消引用 $args 时，每个参数都被拆分为一个单词:

when I unquote the $args, then each argument was split into single word:

hello
world
second
arg
3rdarg

最后，我想知道是否有一种方法可以定义像 $@ 这样的变量.任何回应将不胜感激.

Lastly, I wonder if there's a way to define variable like $@. Any response will be appreciated.

推荐答案

你需要使用一个数组:

function f() {
    local args=("$@")             # values in parentheses
    for arg in "${args[@]}"; do   # array expansion syntax
        echo "$arg"
    done
}

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