Consider:
gndlp@ubuntu:~$ test -x examples.desktop && echo $? gndlp@ubuntu:~$ test -x examples.desktop & echo $? [1] 2992 0 Why is Bash acting the way it is in this situation?
Is the test command simply not finishing and thus the echo command isn't processed?
3 Answers
The meaning of && and & are intrinsically different.
- What is
&&in Bash? In Bash—and many other programming languages—&&means “AND”. And in command execution context like this, it means items to the left as well as right of&&should be run in sequence in this case. - What is
&in Bash? And a single&means that the preceding commands—to the immediate left of the&—should simply be run in the background.
So looking at your example:
gndlp@ubuntu:~$ 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.
& 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.
Look at what your commands are:
test -x examples.desktop && echo $? This means check to see if examples.desktop is executable and if it is then execute echo $?.
test -x examples.desktop & echo $? means check to see if examples.desktop is executable in the "background". Then execute echo $?.
