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in reply to Re: The basics
in thread The basics

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

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(OT) Re^3: The basics
by Hue-Bond (Priest) on Jun 24, 2006 at 21:59 UTC
    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

    --
    David Serrano

      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.

        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.

        --
        David Serrano

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

      - tye