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What does $!# mean in Shell scripting? [duplicate]

How does $!# work in bash to get the last command-line argument?Functional shell scriptingCounters in shellShell Script Size 57Mcronjob bash script called within another shell script not workingWait for key in shell script that may get piped to /bin/bashwhat could be the reasons for shell scripts kill itself?Shell script work on terminal, not when it has run by cronjobCan we write a script to run some command within each interactive bash shell process?How can I skip the rest of a script without exiting the invoking shell, when sourcing the script?What exit modes exist in shell-scripting in general and in Bash in particular?

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26

3

This question already has an answer here:

• 
  How does $!# work in bash to get the last command-line argument?
  
  1 answer
  
  

Found this in an obfuscated malicious shell script, beginning with:

$!#$*^ <<<...

Could not find any reference to $!#, but when echo'd, it outputs -bash. Is this a secret referene to the running shell? Why there is an extra dash then?

Thanks!

shell-script

share|improve this question

asked May 23 at 2:35

DaiDai

23316

New contributor

Dai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

marked as duplicate by Jesse_b, 200_success, X Tian, garethTheRed, Philip Couling May 24 at 14:39

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.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  In "shell scripting", that will expand to the PID of the last background command. In bash (and only in bash), $!# is the last positional parameter obtained via "Indirect Expansion" (the $!var syntax) of the the $# variable.
  
  – mosvy
  May 23 at 14:07
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  @mosvy That looks like an answer, not a clarification of the question. As you point out, it includes information missing from the current answer, which is exactly why the system allows multiple answers to one question.
  
  – IMSoP
  May 23 at 14:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Regarding the extra dash itself, refer to this answer: superuser.com/a/278865/560377
  
  – Phlarx
  May 23 at 16:27
  
  

  
  
  
  

add a comment | 

26

3

This question already has an answer here:

• 
  How does $!# work in bash to get the last command-line argument?
  
  1 answer
  
  

Found this in an obfuscated malicious shell script, beginning with:

$!#$*^ <<<...

Could not find any reference to $!#, but when echo'd, it outputs -bash. Is this a secret referene to the running shell? Why there is an extra dash then?

Thanks!

shell-script

share|improve this question

asked May 23 at 2:35

DaiDai

23316

New contributor

Dai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

marked as duplicate by Jesse_b, 200_success, X Tian, garethTheRed, Philip Couling May 24 at 14:39

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.

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  In "shell scripting", that will expand to the PID of the last background command. In bash (and only in bash), $!# is the last positional parameter obtained via "Indirect Expansion" (the $!var syntax) of the the $# variable.
  
  – mosvy
  May 23 at 14:07
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  @mosvy That looks like an answer, not a clarification of the question. As you point out, it includes information missing from the current answer, which is exactly why the system allows multiple answers to one question.
  
  – IMSoP
  May 23 at 14:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Regarding the extra dash itself, refer to this answer: superuser.com/a/278865/560377
  
  – Phlarx
  May 23 at 16:27
  
  

  
  
  
  

add a comment | 

This question already has an answer here:

• 
  How does $!# work in bash to get the last command-line argument?
  
  1 answer
  
  

Found this in an obfuscated malicious shell script, beginning with:

$!#$*^ <<<...

Could not find any reference to $!#, but when echo'd, it outputs -bash. Is this a secret referene to the running shell? Why there is an extra dash then?

Thanks!

shell-script

share|improve this question

asked May 23 at 2:35

DaiDai

23316

New contributor

Dai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked May 23 at 2:35

DaiDai

23316

New contributor

Dai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked May 23 at 2:35

DaiDai

23316

New contributor

Dai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 23 at 2:35

DaiDai

23316

New contributor

Dai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 23 at 2:35

DaiDai

23316

1 Answer
1

active

oldest

votes

35

This answer is bash-specific, because you tried the echo in bash. Not all shells behave the same.

In bash, if $var1 is foo, then $!var1 is the same as $foo. The ! is indirect expansion: it causes bash to retrieve from the given variable a variable name (foo) instead of a value.

Now replace var1 with #. 
$# is the number of arguments. 
If $# is 0, $!# is $0. 
If $# is 4, $!# is $4. 
In other words, $!# is the last positional parameter, no matter how many there positional parameters there are.

If there are no positional parameters, $# is 0, so the result is $0, which is the name of the shell or script (reference). In your case, that was bash (plus a leading - meaning that it's a login shell).

Quick test:

$ echo $!#
-bash

$ set the quick brown fox

$ echo $!#
fox

share|improve this answer

edited May 24 at 20:01

cxw

95111013

answered May 23 at 2:51

G-ManG-Man

14.5k94174

• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  
  Your answer only applies to bash, but you fail to mention that. In all the other shells, $!# will expand to the PID of the last background command ($!) with a empty string removed from the beginning (the $var#pat form of parameter expansions).
  
  – mosvy
  May 23 at 13:55
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Comments are indeed for clarifying answers; while it's possible to assume bash, given the text in the question, it's certainly reasonable to (collaboratively!) edit this answer to indicate shell-specific behavior.
  
  – Jeff Schaller♦
  May 23 at 17:58
  

  
  
  
  

add a comment | 

35

This answer is bash-specific, because you tried the echo in bash. Not all shells behave the same.

In bash, if $var1 is foo, then $!var1 is the same as $foo. The ! is indirect expansion: it causes bash to retrieve from the given variable a variable name (foo) instead of a value.

Now replace var1 with #. 
$# is the number of arguments. 
If $# is 0, $!# is $0. 
If $# is 4, $!# is $4. 
In other words, $!# is the last positional parameter, no matter how many there positional parameters there are.

If there are no positional parameters, $# is 0, so the result is $0, which is the name of the shell or script (reference). In your case, that was bash (plus a leading - meaning that it's a login shell).

Quick test:

$ echo $!#
-bash

$ set the quick brown fox

$ echo $!#
fox

share|improve this answer

edited May 24 at 20:01

cxw

95111013

answered May 23 at 2:51

G-ManG-Man

14.5k94174

• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  
  Your answer only applies to bash, but you fail to mention that. In all the other shells, $!# will expand to the PID of the last background command ($!) with a empty string removed from the beginning (the $var#pat form of parameter expansions).
  
  – mosvy
  May 23 at 13:55
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Comments are indeed for clarifying answers; while it's possible to assume bash, given the text in the question, it's certainly reasonable to (collaboratively!) edit this answer to indicate shell-specific behavior.
  
  – Jeff Schaller♦
  May 23 at 17:58
  

  
  
  
  

add a comment | 

35

This answer is bash-specific, because you tried the echo in bash. Not all shells behave the same.

In bash, if $var1 is foo, then $!var1 is the same as $foo. The ! is indirect expansion: it causes bash to retrieve from the given variable a variable name (foo) instead of a value.

Now replace var1 with #. 
$# is the number of arguments. 
If $# is 0, $!# is $0. 
If $# is 4, $!# is $4. 
In other words, $!# is the last positional parameter, no matter how many there positional parameters there are.

If there are no positional parameters, $# is 0, so the result is $0, which is the name of the shell or script (reference). In your case, that was bash (plus a leading - meaning that it's a login shell).

Quick test:

$ echo $!#
-bash

$ set the quick brown fox

$ echo $!#
fox

share|improve this answer

edited May 24 at 20:01

cxw

95111013

answered May 23 at 2:51

G-ManG-Man

14.5k94174

• 
  
  
  

  
  19
  

  
  
  
  
  
  

  
  
  Your answer only applies to bash, but you fail to mention that. In all the other shells, $!# will expand to the PID of the last background command ($!) with a empty string removed from the beginning (the $var#pat form of parameter expansions).
  
  – mosvy
  May 23 at 13:55
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Comments are indeed for clarifying answers; while it's possible to assume bash, given the text in the question, it's certainly reasonable to (collaboratively!) edit this answer to indicate shell-specific behavior.
  
  – Jeff Schaller♦
  May 23 at 17:58
  

  
  
  
  

add a comment | 

This answer is bash-specific, because you tried the echo in bash. Not all shells behave the same.

In bash, if $var1 is foo, then $!var1 is the same as $foo. The ! is indirect expansion: it causes bash to retrieve from the given variable a variable name (foo) instead of a value.

Now replace var1 with #. 
$# is the number of arguments. 
If $# is 0, $!# is $0. 
If $# is 4, $!# is $4. 
In other words, $!# is the last positional parameter, no matter how many there positional parameters there are.

If there are no positional parameters, $# is 0, so the result is $0, which is the name of the shell or script (reference). In your case, that was bash (plus a leading - meaning that it's a login shell).

Quick test:

$ echo $!#
-bash

$ set the quick brown fox

$ echo $!#
fox

share|improve this answer

edited May 24 at 20:01

cxw

95111013

answered May 23 at 2:51

G-ManG-Man

14.5k94174

$ echo $!#
-bash

$ set the quick brown fox

$ echo $!#
fox

share|improve this answer

edited May 24 at 20:01

cxw

95111013

answered May 23 at 2:51

G-ManG-Man

14.5k94174

edited May 24 at 20:01

cxw

95111013

edited May 24 at 20:01

cxw

95111013

answered May 23 at 2:51

G-ManG-Man

14.5k94174

answered May 23 at 2:51

G-ManG-Man

14.5k94174