bash-为什么（cd && sleep 5000）开始第二个进程

Question

I have this script test.sh

#!/bin/bash
( cd . && sleep 5000 )

I execute with ./test.sh & and then run ps lax | grep test.sh ps lax | grep test.sh

I now have 2 processes running...

0  1000  6883  6600  20   0  10600  1332 -      S    pts/2      0:00 /bin/bash ./test.sh
1  1000  6884  6883  20   0  10604   704 -      S    pts/2      0:00 /bin/bash ./test.sh

• Why do I have two processes running and where does the second process come from?
• Why don't I have two processes if I remove the cd ".." from the command?

Thanks for any explanation, I just don't get it and I think I'm lacking some basics here... or is this some vodoo? ;)

Answer 1

在括号内对一系列bash命令进行分组将在子shell中执行它们。

Answer 2

Writing two commands together might pose some difficulties for Terminal to detect the commands. Perhaps you could split it up into 2 commands.

cd ...

sleep 5000