Bash – What is `$?` in Bash?

bashcommand line

The question is pretty straight forward. I can get the exit code of the previous command by using $?. But what exactly is it?

I don't think it is a variable, because you can't have any special characters in the name of variable other than _.

foo?foo=10

will result in the error: foo?foo=10: command not found

So if it is not a variable, what is it? Are there others like it?

Best Answer

What is it?

$? is a built-in variable that stores the exit status of a command, function, or the script itself.

$? reads the exit status of the last command executed. After a function returns, $? gives the exit status of the last command executed in the function. This is Bash's way of giving functions a "return value." It returns 0 on success or an integer in the range 1 - 255 on error.

Are there others like it too?

Yes,there are several such built-in variables in bash. You can see a list here. Refer: http://www.tldp.org/LDP/abs/html/exit-status.html

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BASH_COMMAND Variable – What is it Used For?

Answering to the third question: of course it can be used meaningfully in the way at Bash manual clearly hints – in a trap, e. g.:

$ trap 'echo ‘$BASH_COMMAND’ failed with error code $?' ERR
$ fgfdjsa
fgfdjsa: command not found
‘fgfdjsa’ failed with error code 127
$ cat /etc/fgfdjsa
cat: /etc/fgfdjsa: No such file or directory
‘cat /etc/fgfdjsa’ failed with error code 1

Ubuntu – Execute a command stored into a $variable

Instead of using simple strings, build your command using arrays. Arrays provide a convenient interface: If a is an array, then "${a[@]}" (note the quotes) expands into each element of a, without additional field splitting or globbing (so spaces, and things like wildcards, should remain intact).

Example:

$ a=(printf "|%s|\n" "foo   bar" "*")
$ echo ${a[@]}
printf |%s|\n foo bar bin boot dev etc home lib lib64 lost+found mnt opt proc root run sbin srv sys tmp usr var

Note how the * was expanded, and how the extra spaces between foo and bar were lost. But with the "${a[@]}", these are preserved:

$ echo "${a[@]}"
printf |%s|\n foo   bar *

This is ideal for building commands. Now, you can do:

$ "${a[@]}"
|foo   bar|
|*|

See? The arguments were retained perfectly.

So, do:

COMMAND=(/usr/bin/exiftool "$PATH_NAME")
"${COMMAND[@]}"

Best Answer

Related Solutions

BASH_COMMAND Variable – What is it Used For?

Ubuntu – Execute a command stored into a $variable

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