Programming Forums > Script Programming > Bash / Shell Scripting

Jimbo · Nov 16th, 2006, 3:47 AM

Someone mentioned to me recently that the ! part of the #! to start a shell script had some special meaning. He mentioned something along the lines of how the # denotes a comment, but the ! causes the line to still be interpreted somehow. So I was playing around with it and noticed some odd things. It seems like the script calls itself, but with itself as a parameter? Or does the #! line just prepend itself to the list of parameters actually passed in (hence how a script gets itself as $0)?

First some oddities with echo:

(Toggle Plain Text)

jimmy@vera ~/code/bash $ cat blah.sh 
#!/bin/echo Hello 
echo world
jimmy@vera ~/code/bash $ ./blah.sh 
Hello ./blah.sh

I can understand if it were to execute either the second line (with the first ignored as a comment), or the second (and stopping after running echo). But it echo's both Hello and the filename...

Similarly with cat:

(Toggle Plain Text)

jimmy@vera ~/code/bash $ cat me
this file is called me
jimmy@vera ~/code/bash $ cat blah.sh 
#! /bin/cat me 
echo Hello world
jimmy@vera ~/code/bash $ ./blah.sh 
this file is called me
#! /bin/cat me 
echo Hello world

Again, it executes the first line (with the #!), but not the second, and again it adds itself as a parameter to the first line.

and to be neat and tidy, I tried rm as well:

(Toggle Plain Text)

jimmy@vera ~/code/bash $ cat blah.sh 
#! /bin/rm me 
echo hello world
jimmy@vera ~/code/bash $ ls
blah.sh  filenamesWithSpaces.sh  me
jimmy@vera ~/code/bash $ ./blah.sh 
jimmy@vera ~/code/bash $ ls
filenamesWithSpaces.sh

The second guess I came up with seems almost logical, except that at this point I'd modify it to say it somehow passes the file and its arguments to whatever is cited after the #!. But I'm still making guesses, so does anyone know how it actually works?

Arevos · Nov 16th, 2006, 4:01 AM

It's called a shebang. It's the *nix way of determining what interpreter to use when executing a shell script. The operating system looks for the shebang at the start of the script its executing, and then passes the entire contents of the file to the program specified by the shebang. Any additional arguments are appended to the end of the program.

For instance, say I have a file called hello.fb:

(Toggle Plain Text)

#! /usr/bin/foobar -y
All of this text as well as the line above
is passed to "foobar -y" as STDIN

Then this line:

(Toggle Plain Text)

./hello.fb world

Is the same as:

(Toggle Plain Text)

/usr/bin/foobar -y world < hello.fb

DaWei · Nov 16th, 2006, 7:08 AM

Now for the useless information: shebang is a corruption of hash bang, the # being the hash mark and the ! being the bang mark.

jim mcnamara · Nov 16th, 2006, 5:27 PM

It constitutes a magic number. UNIX can interpret the magic number for a lot of files, and then behave accordingly.

#!/bin/ksh
loads the ksh shell for the duration of the script - the script file becomes stdin.

look for /usr/include/magic.h and /usr/include/sys/magic.h for more information. man magic or info magic may also help.

mackenga · Dec 1st, 2006, 5:47 PM

I've heard a few UK-based Unix hackers call it the "Whiz bang." Not really very informative, I know, but I find it quite amusing. Maybe I should get out more...

Nov 16th, 2006, 3:47 AM	#1
Jimbo Battle Programmer Join Date: Feb 2006 Location: Bellevue, WA, USA Posts: 770 Rep Power: 3	What does the ! do in #! at the start of a .sh? Someone mentioned to me recently that the ! part of the #! to start a shell script had some special meaning. He mentioned something along the lines of how the # denotes a comment, but the ! causes the line to still be interpreted somehow. So I was playing around with it and noticed some odd things. It seems like the script calls itself, but with itself as a parameter? Or does the #! line just prepend itself to the list of parameters actually passed in (hence how a script gets itself as $0)? First some oddities with echo: (Toggle Plain Text) jimmy@vera ~/code/bash $ cat blah.sh #!/bin/echo Hello echo world jimmy@vera ~/code/bash $ ./blah.sh Hello ./blah.sh jimmy@vera ~/code/bash $ cat blah.sh #!/bin/echo Hello echo world jimmy@vera ~/code/bash $ ./blah.sh Hello ./blah.sh I can understand if it were to execute either the second line (with the first ignored as a comment), or the second (and stopping after running echo). But it echo's both Hello and the filename... Similarly with cat: (Toggle Plain Text) jimmy@vera ~/code/bash $ cat me this file is called me jimmy@vera ~/code/bash $ cat blah.sh #! /bin/cat me echo Hello world jimmy@vera ~/code/bash $ ./blah.sh this file is called me #! /bin/cat me echo Hello world jimmy@vera ~/code/bash $ cat me this file is called me jimmy@vera ~/code/bash $ cat blah.sh #! /bin/cat me echo Hello world jimmy@vera ~/code/bash $ ./blah.sh this file is called me #! /bin/cat me echo Hello world Again, it executes the first line (with the #!), but not the second, and again it adds itself as a parameter to the first line. and to be neat and tidy, I tried rm as well: (Toggle Plain Text) jimmy@vera ~/code/bash $ cat blah.sh #! /bin/rm me echo hello world jimmy@vera ~/code/bash $ ls blah.sh filenamesWithSpaces.sh me jimmy@vera ~/code/bash $ ./blah.sh jimmy@vera ~/code/bash $ ls filenamesWithSpaces.sh jimmy@vera ~/code/bash $ cat blah.sh #! /bin/rm me echo hello world jimmy@vera ~/code/bash $ ls blah.sh filenamesWithSpaces.sh me jimmy@vera ~/code/bash $ ./blah.sh jimmy@vera ~/code/bash $ ls filenamesWithSpaces.sh The second guess I came up with seems almost logical, except that at this point I'd modify it to say it somehow passes the file and its arguments to whatever is cited after the #!. But I'm still making guesses, so does anyone know how it actually works? __________________ <insert disclaimer here> <insert shameless plug for Visual Studio here>

Nov 16th, 2006, 4:01 AM	#2
Arevos Programming Guru Join Date: Aug 2005 Location: England Posts: 1,499 Rep Power: 5	It's called a shebang. It's the *nix way of determining what interpreter to use when executing a shell script. The operating system looks for the shebang at the start of the script its executing, and then passes the entire contents of the file to the program specified by the shebang. Any additional arguments are appended to the end of the program. For instance, say I have a file called hello.fb: (Toggle Plain Text) #! /usr/bin/foobar -y All of this text as well as the line above is passed to "foobar -y" as STDIN #! /usr/bin/foobar -y All of this text as well as the line above is passed to "foobar -y" as STDIN Then this line: (Toggle Plain Text) ./hello.fb world ./hello.fb world Is the same as: (Toggle Plain Text) /usr/bin/foobar -y world < hello.fb /usr/bin/foobar -y world < hello.fb

Nov 16th, 2006, 7:08 AM	#3
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Now for the useless information: shebang is a corruption of hash bang, the # being the hash mark and the ! being the bang mark. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Dec 1st, 2006, 5:47 PM	#5
mackenga Professional Programmer Join Date: Mar 2005 Location: Glasgow, Scotland Posts: 328 Rep Power: 4	I've heard a few UK-based Unix hackers call it the "Whiz bang." Not really very informative, I know, but I find it quite amusing. Maybe I should get out more... __________________ "I'm not a genius. Why do I have to suffer?"

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Nov 16th, 2006, 5:27 PM	#4
jim mcnamara Hobbyist Programmer Join Date: Jun 2005 Location: New Mexico Posts: 228 Rep Power: 4	It constitutes a magic number. UNIX can interpret the magic number for a lot of files, and then behave accordingly. #!/bin/ksh loads the ksh shell for the duration of the script - the script file becomes stdin. look for /usr/include/magic.h and /usr/include/sys/magic.h for more information. man magic or info magic may also help.

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Programming Forums > Script Programming > Bash / Shell Scripting

What does the ! do in #! at the start of a .sh?