在sh中初始化数组 [英] initializing arrays in sh

问题描述

我想在sh中初始化一个数组.

I want to initialize an array in sh.

用bash表示:

list=(`seq 1 4`)

在sh中，我尝试这样做:

In sh I try to do it like this:

    for i in `seq 1 4`; do
        list[$((i-1))]="$i"
    done

尽管每次迭代我都会出错:

I get an error though for each iteration saying:

list[0]=1: not found

我在做什么错以及如何解决?

What am I doing wrong and how to fix that?

推荐答案

POSIX sh 不支持数组.您需要一个更高级的外壳程序，例如bash，zsh或ksh.

POSIX sh doesn't support arrays. You need a more advanced shell for that, e.g. bash, zsh, or ksh.

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