Does anyone know why the following script works?
#a-random-junk-string
echo HI
The shell executes the echo command, and outputs HI. I thought that since there is no "!" after the "#", the shell would give an error.
+12
A:
If there is no #! specifying a specific interpreter, the kernel will not intercept and launch it with the specified program.
However, the current shell may still interpret it as a command file, which is what you are seeing take place.
Amardeep
2010-07-26 17:16:52
This is correct. The reason the script still works is that '#' is the comment character in Bash/SH/ZSH/etc, so the first line is ignored as a comment.
Borealid
2010-07-26 17:18:29
Thanks for the complete answer!
Yassin
2010-07-26 17:18:29
+1
A:
When the shell is asked to run a file with the executable bit turned on then it will examine the file and determine if it begins with a shebang #! if it does then it will execute that command which will get it's program text from the remainder of the file.
If the file does not start with a shebang then the shell will attempt to execute it itself. This is what is happening for you and the shell interprets the first line as a comment.
Steve Weet
2010-07-26 17:21:16
Actually, the loader does this, not the shell. See http://www.in-ulm.de/~mascheck/various/shebang/sys1.c.html for Dennis Ritchie's announcement of this feature when it was added to Unix. It gives a wonderful explanation of why it's better for the loader to do it than for the shell to do it.
Dan Moulding
2010-07-26 17:39:44