'&&' vs. '&' with the 'test' command in Bash

Question

Consider:

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

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

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

Answer 1

The meaning of && and & are intrinsically different.

• 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.

So looking at your example:

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

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.

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.

Answer 2

& executes a command in the background, and will return 0 regardless of its status.

From the man page:

If a command is terminated by the control operator &, the shell executes the command in the background in a subshell. The shell does not wait for the command to finish, and the return status is 0. Commands separated by a ; are executed sequentially; the shell waits for each command to terminate in turn. The return status is the exit status of the last command executed.