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8735Re: "ocaml_beginners"::[] Choosing the order of parameters

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  • Peng Zang
    Oct 1, 2007
      Hash: SHA1

      Ahh, right. So in func2 the function is explictly defined in two parts and so
      there is actual partial application. I thought about why ocaml doesn't do
      this by default but it makes sense as then the timing of any side-effects in
      your functions would be ambiguous. Thanks for the example,


      On Monday 01 October 2007 02:35, dmitry grebeniuk wrote:
      > Hello, Peng.
      > PZ> Apropo partial application, I didn't think ocaml
      > PZ> did partial application... I mean you can always
      > PZ> give a function fewer arguments than it takes and
      > PZ> ocaml will return a function with the remaining
      > PZ> arguments but I didn't think ocaml did any thing
      > PZ> (ie. evaluation/calculation) with the arguments
      > PZ> that it has gotten. I thought it just held the
      > PZ> arguments it has until it has all the arguments
      > PZ> at which point, the function is then evaluated.
      > That's not totally right. Try to run the following
      > code and get the difference between partial applications:
      > ======================================================
      > let func1 () () =
      > print_string "func1: fully applied\n";;
      > let func2 () =
      > let () = print_string "func2: partially applied\n" in
      > fun () -> print_string "func2: fully applied\n";;
      > let _ =
      > print_string "1\n";
      > let p1 = func1 () in
      > print_string "2\n";
      > p1 ();
      > print_string "3\n";
      > let p2 = func2 () in
      > print_string "4\n";
      > p2 ();
      > print_string "5\n";;
      > ======================================================
      > Sometimes this difference is useful, it's like a
      > "poor men's OOP": you can do some init-stuff when
      > partially applying first argument(s), and then use
      > resulting closure.
      > But it's not a good style of programming, imho.
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