如何阅读shell命令的源代码? [英] How do I read the source code of shell commands?

问题描述

我想阅读用linux命令编写的实际源代码.我已经获得了使用它们的一些经验，现在我认为是时候与我的机器进行更深层次的互动了.

I would like to read the actual source code which the linux commands are written with. I've gained some experience using them and now I think it's time to interact with my machine at a deeper level.

我在这里找到了一些命令 http://directory.fsf.org/wiki/GNU.不幸的是，我找不到基本的命令，例如"ls"，在我看来这很容易上手.

I've found some commands here http://directory.fsf.org/wiki/GNU. Unfortunately I wasn't able to find basic commands such as 'ls' which seems to me easy enough to begin.

我如何准确地读取诸如'ls'之类的简单shell命令的源代码? 我在Ubuntu 12.04上运行

How exactly do I read the source code of the simple shell commands like 'ls'? I'm running on Ubuntu 12.04

推荐答案

所有这些基本命令都是coreutils软件包的一部分.

All these basic commands are part of the coreutils package.

您可以在此处找到所需的所有信息:

You can find all information you need here:

http://www.gnu.org/software/coreutils/

如果要下载最新的源代码，则应使用git:

If you want to download the latest source, you should use git:

git clone git://git.sv.gnu.org/coreutils

要在Ubuntu计算机上安装git，应使用apt-get(标准Ubuntu安装中不包含git):

To install git on your Ubuntu machine, you should use apt-get (git is not included in the standard Ubuntu installation):

sudo apt-get install git

需要说明的是，在这里您可以找到ls命令的特定来源:

Truth to be told, here you can find specific source for the ls command:

http://git.savannah.gnu. org/cgit/coreutils.git/tree/src/ls.c

仅4984条代码行足以像ls一样使命令足够简单" ...您仍然对阅读它感兴趣吗?祝你好运！ :D

Only 4984 code lines for a command 'easy enough' as ls... are you still interested in reading it?? Good luck! :D

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