Programming Forums - View Single Post - What does the ! do in #! at the start of a .sh?

Arevos · Nov 16th, 2006, 3:01 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:

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

Then this line:

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./hello.fb world

Is the same as:

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

Nov 16th, 2006, 3:01 AM	#2
Arevos Programming Guru Join Date: Aug 2005 Location: England Posts: 1,499 Rep Power: 4	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