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This question already has an answer here:

I've always used !$ to refer to the last argument of the previous command.

e.g.

~/dir » mkdir birthday
~/dir » cd !$
~/dir » cd birthday

However I've started seeing tutorials use $_ in place of !$

I've tested the two and the only difference I've noticed is subtle, but it appears that $_ will evaluate and execute the command whereas !$ will let you sanity check the output before hitting Enter, but is this the only difference between $_ and !$?

Update: flagged as a duplicate? This question may be answered by the accepted answer for the suggested question, but only because the answer is more detailed than the question required.

Searching for What does _$ mean? does yield the suggested duplicate, however I already and I knew prior to asking what !$ did, and there is nothing that suggests that question might also answer other questions.

marked as duplicate by αғsнιη, Stephen Rauch, Romeo Ninov, Archemar, Wieland Nov 4 '17 at 9:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

7

man bash, search for section Special Parameters. For _ you'll find:

expands to the last argument to the previous command, after expansion.

So in a command line context the difference is the expansion. For example if I have a variable FOO set to 'bar' and I execute echo $FOO then $_ on the next line will result in bar being submitted in the next command while !$ will result in $FOO.

This behavior might not be obvious by default but with shopt -s histverify it's a little more evident. (With histverify enabled Bash will not submit history expansion immediately after Enter, it will first display the command, post-expansion, and wait for a second Enter). So we would see something like this after FOO=bar:

$ echo $FOO
bar
$ ls !$ $_
# We hit Enter and then see
$ ls $FOO $_ 

$_ is not expanded until the command is submitted during parameter expansion (along with $FOO). It's a subtle distinction but there you go.

Probably the more interesting/useful distinction is that $_ is a variable. So, for example, if the last parameter were /some/file/path.txt you could get it's length ${#_} -> 19, extract filename only ${_##*/} -> path.txt, change the root ${_//some/other} -> /other/file/path.txt and so on. Can't do all of that with !$.


Note that $_ has different meanings in other contexts:

At shell startup, set to the absolute pathname used to invoke the shell or shell script being executed as passed in the environment or argument list.

...and...

Also set to the full pathname used to invoke each command executed and placed in the environment exported to that command.

...and...

When checking mail, this parameter holds the name of the mail file currently being checked.


UPDATE: While mostly accurate, my answer is lacking some key information and perhaps even misleading about how/when $_ actually gets a value. That first quote is meant to read like this:

expands to the last argument to the previous command as it was received by that command, i.e. after going through Shell Expansion

Best way to illustrate is with a couple examples:

$ echo "$(date +%s)" # %s is seconds since epoch
1536984661
$ echo $_  "$(date +%s)" # don't hit enter until a few seconds pass
1536984661 1536984665

So $_ gets the actual seconds-since-epoch calculated at the time the first echo executed not something closer to the 1536984665 of the second date invocation.

Another one:

$ foo='a b c'
$ echo "$foo"
a b c
$ echo "$_"
a b c
$ #---------
$ echo $foo
a b c
$ echo "$_"
c

In the second case we didn't quote $foo. Before being passed to echo it went through variable/parameter expansion followed by word splitting. The lack of quotes resulted in "a", "b" and "c" being passed as three separate parameters and $_ got the last, i.e. "c".

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