'&&'与“&"在Bash中使用'test'命令 [英] '&&' vs. '&' with the 'test' command in Bash

问题描述

考虑:

gndlp@ubuntu:~$ test -x examples.desktop  && echo $?
gndlp@ubuntu:~$ test -x examples.desktop  & echo $?
[1] 2992
0

在这种情况下，为什么Bash会表现出来?

Why is Bash acting the way it is in this situation?

测试命令是否只是没有完成，因此没有处理echo命令?

Is the test command simply not finishing and thus the echo command isn't processed?

推荐答案

&&和&的含义本质上是不同的.

The meaning of && and & are intrinsically different.

• Bash中的&&是什么??在Bash和许多其他编程语言中，&&的意思是"AND".并且在这样的命令执行上下文中，这意味着在这种情况下，应按顺序运行&&左侧和右侧的项目.
• Bash中的&是什么??单个&意味着前面的命令(在&的最左端)应该只在后台运行.

• What is && in Bash? In Bash—and many other programming languages—&& means "AND". And in command execution context like this, it means items to the left as well as right of && should be run in sequence in this case.
• What is & in Bash? And a single & means that the preceding commands—to the immediate left of the &—should simply be run in the background.

因此，请看您的示例:

gndlp@ubuntu:~$ test -x examples.desktop  && echo $?
gndlp@ubuntu:~$ test -x examples.desktop  & echo $?
[1] 2992
0

第一个命令(实际上是结构化的)实际上不返回任何内容.但是第二条命令返回[1] 2992，其中2992引用在后台运行的进程ID(PID)，而0是第一条命令的输出.

The first command—as it is structured—actually does not return anything. But second command returns a [1] 2992 in which the 2992 refers to the process ID (PID) that is running in the background and the 0 is the output of the first command.

由于第二个命令仅在后台运行test -x examples.desktop，因此它很快发生，因此生成了进程ID，并立即消失了.

Since the second command is just running test -x examples.desktop in the background it happens quite quickly, so the process ID is spawned and gone pretty immediately.

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