What does the line "#!/bin/sh" mean in a UNIX shell script?

Question

It's called a shebang, and tells the parent shell which interpreter should be used to execute the script.

#!/bin/sh <--------- bourne shell compatible script
#!/usr/bin/perl  <-- perl script
#!/usr/bin/php  <--- php script
#!/bin/false <------ do-nothing script, because false returns immediately anyways.

Most scripting languages tend to interpret a line starting with # as comment and will ignore the following !/usr/bin/whatever portion, which might otherwise cause a syntax error in the interpreted language.

When you try to execute a program in unix (one with the executable bit set), the operating system will look at the first few bytes of the file. These form the so-called "magic number", which can be used to decide the format of the program and how to execute it.

#! corresponds to the magic number 0x2321 (look it up in an ascii table). When the system sees that the magic number, it knows that it is dealing with a text script and reads until the next \n (there is a limit, but it escapes me atm). Having identified the interpreter (the first argument after the shebang) it will call the interpreter.

Other files also have magic numbers. Try looking at a bitmap (.BMP) file via less and you will see the first two characters are BM. This magic number denotes that the file is indeed a bitmap.

If the file that this script lives in is executable, the hash-bang (#!) tells the operating system what interpreter to use to run the script. In this case it's /bin/sh, for example.

There's a Wikipedia article about it for more information.

The first line tells the shell that if you execute the script directly (./run.sh; as opposed to /bin/sh run.sh), it should use that program (/bin/sh in this case) to interpret it.

You can also use it to pass arguments, commonly -e (exit on error), or use other programs (/bin/awk, /usr/bin/perl, etc).