Re^2: The basics

[Joost]

The #! line has two programs that "interpret" it:

• To the shell. An executable file that starts with a #! line will cause the executing shell to start up the program indicated in that line. That way, you can use #!/bin/sh to create a shell script, #!/usr/bin/perl to create a perl script etc. At least on unixish systems, there is always a shell involved in this process (but for instance windows doesn't work like that)

• (Possibly) to the interpreter indicated in the #! line.

  For instance, perl reads the additional arguments on the #! line as a sort of "setup parameters". Example: #!/usr/bin/perl -wT will run perl with global warnings enabled and in taint mode.

[Hue-Bond]

To the shell. [...] At least on unixish systems, there is always a shell involved in this process

The #! line, also called shebang, tells nothing to any shell. It's the kernel who interprets it. When the first two bytes of an executable file are #!, the kernel runs the command that follows them, with the parameters specified, and feeds it the whole file on its standard input:

$ cat > a
#!/usr/bin/tail -n2
1
2
3
4
5
6
7
^D
$ chmod +x a
$ ./a
6
7
$ strace ./a 2>&1 | grep -c sh
0

[tye]

Just FYI: That is only true for recent Unix versions. Originally, the #! line was interpretted (only) by the shell(s).

[Tanktalus]

I'm trying to figure out where you get the idea that the contents are passed to tail's stdin. It's not. The full path to the file (as given by the shell - full path if searched via PATH, relative path if PATH not used) is passed in.

Compare the output of "strace ./a" and "strace /usr/bin/tail -n2 < a". The former has a call to "open" and all reading is done from filehandle 3. The latter doesn't call open, and all reading is done from filehandle 0.

[Hue-Bond]

I'm trying to figure out where you get the idea that the contents are passed to tail's stdin.

I guess I'll also have to. To my surprise, you're right.