Unix $#命令有什么作用? [英] What does the unix $# command do?

问题描述

我阅读了此页面什么是特殊的美元符号外壳变量?，但我仍然不明白 $#的作用.

I read this page What are the special dollar sign shell variables? but I still don't get what $# does.

我有一个演讲幻灯片的例子:

I have an example from a lecture slide:

#!/bin/sh
echo Total no. of inputs: $#
echo first: $1
echo second: $2

我假设$#接受所有输入作为参数，我们应该期望有两个输入.那是在做什么吗?

I assume $# takes in all inputs as the argument, and we should expect two inputs. Is that what it's doing?

推荐答案

正如您的脚本所说:您要传递给脚本的命令行参数总数.

as your script says: Its total number of command line arguments you are passing to your script.

如果您的脚本名称为: kl.sh

if you have a script name as: kl.sh

执行为

./kl.sh hj jl lk 甚至 bash kl.sh hj jl lk

在脚本中您正在做

echo $#

它将打印 3

其中

$1 is hj
$2 is jl
$3 is lk

本教程肯定会为您提供帮助

This tutorial will surely help you

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