• ronigami@lemmy.world
      11·
      6 days ago

      That’s only one advantage. In theory it does not necessarily terminate, so that’s another one.

  • ulterno@programming.devEnglish
    24·
    6 days ago

    That’s not even enough to get you a job these days.
    You now have to use:

    do {
    x = reinterpret_cast<int>(AI::Instance().ask("Do Something. Anything. Be efficient and productive. Use 10 tokens."));
} while (x != 10);
    • melfie@lemy.lol
      6·
      6 days ago

      You’re absolutely right! Who sets a variable these days without running it though a LLM?

        • MonkeMischief@lemmy.today
          2·
          5 days ago

          First, we’ll deep dive into “What is a variable?”, then together we’ll examine “Who sets a variable?”, “What is an LLM?” and finally, “Who would set a variable without using an LLM?”

          You’ll be a coding pro in no time!

          How does that sound?

          (I felt gross writing this lmao)

    • okmko@lemmy.world
      7·
      7 days ago

      Freshman year of college doing assembly programming, I spent a while figuring out a “programmic” way to solve a problem, trying to wrangle labels and gotos. My friend came in with essentially this but as lookup table. It blew my mind.

      It was then that I learned to trade space for complexity.

    • TheOakTree@lemmy.zip
      2·
      6 days ago

      If only I could measure the quality of my paper purely by word count…

      I thought “a a a a a a” x100000 was thought-provoking and well tested.

    • cooligula@sh.itjust.worksEnglish
      3·
      6 days ago

      I’d say Meta hiring someone to work on WhatsApp. Man, is that piece of software crap… Every update, a new UI bug/glitch appears

  • Atlas_@lemmy.world
    8·
    7 days ago

    Oddly enough, out of all of these the one the compiler has the best chance of optimizing out is the last one

    • LeFantome@programming.dev
      2·
      6 days ago

      What?

      First one is optimized obvious.

      Second one optimizes to x = 10 via constant propagation.

      Third one first unrolls the loop, propagates constants including booleans, and then eliminates dead code to arrive at x = 10.

      The last one cannot be optimized as “new” created objects that get used, nextInt() changes the state of those objects, and the global state of the random number system is impacted.

  • untorquer@lemmy.world
    3·
    6 days ago

    Something like

    int *a = new int(10)

Int*b = null

While *b !=10 { b = rand(); a=new int(10)}

Return *b

    I haven’t coded recently enough in c/c++ to remember syntax but the concept might work eventually if you’re lucky and have enough memory… Might need a time variant seed on the rand()…

  • spongebue@lemmy.world
    2·
    7 days ago

    x = -i;

    Do many languages let you do that? When it’s in front of a variable I would’ve expected it to be a subtraction operator only and you would need to do x = -1 * i;

    • EvilHankVenture@lemmy.world
      5·
      7 days ago

      In most languages I’ve seen - is both a unary negation operator and a subtraction operator depending on context. So it would negate an integer literal or a variable in this context.

    • squaresinger@lemmy.world
      42·
      6 days ago

      Why would they not let you do that? I honestly don’t know a single language that wouldn’t let you do that. Same as basic math notation allows you to do that.

      x = -i

      is a totally valid mathematical equation.

      For the downvoters: Find me a single language that supports operators but doesn’t have an unary minus operator

      • spongebue@lemmy.world
        1·
        6 days ago

        It’s a valid mathematical notation, sure. But there is an implicit understanding that the - in this case is making a number negative rather than subtracting (or, an implicit subtraction from 0).

        With the way negative numbers generally work in binary there would be much different ones and zeroes stored behind the scenes, so handling that would have to be pretty intentional.

        That said, I did just try it in Java because that’s what I work in normally and I swear I had a gotcha with that. But it worked fine as far as I can tell.

        • squaresinger@lemmy.world
          2·
          5 days ago

          Find me a language where it doesn’t work like that, and we’ll continue the discussion.

          Unary minus operator is standard in every single language that I used so far, including C/C++, Java, Python, Kotlin, Lua, JS/TS, Groovy, PHP, Visual Basic, Excel, Mathematica, Haskell, Bash.

          Here’s more info btw: https://en.wikipedia.org/wiki/Unary_operation

      • squaresinger@lemmy.world
        3·
        6 days ago

        Nope, it is not.

        x = 5
i = 2
x -= i // x => 3

        while

        x = 5
i = 2
x = -i // x => -2

        x=-i is the unary minus operator which negates the value right of it. It doesn’t matter if that value is a literal (-3), a variable (-i) or a function (-f()).

        x-=i is short for x = x-i, and here it’s a binary subtraction, so x is set to the result of i subtracted from x.

        • quilan@lemmy.world
          2·
          5 days ago

          I need to append /s to my future silly replies I think… that said, I’ll never pooh-pooh a well thought response, so thanks for the nice write-up!

          • squaresinger@lemmy.world
            2·
            5 days ago

            Thanks, I totally missed your sarcasm :)

            There’s a couple people in this threat who seem to actually think that x = -i is some weird magic instead of a standard feature that’s present in every major programming language.

      • spongebue@lemmy.world
        1·
        6 days ago

        That only works if x is already 0

        If i is 10 and x is zero, yes, x -= i would have a value of -10. If x was 5 from something else previously, x-=i would end with an x value of -5.

    • boonhet@sopuli.xyz
      1·
      7 days ago

      Personally I would expect it to behave the same in front of a numeric literal and in front of a variable. I do think most languages do that, but I haven’t actually tested that many and could br wrong.

      • squaresinger@lemmy.world
        1·
        5 days ago

        Pretty much all languages do that. It’s a very basic language feature inherited from basic maths notation. Same as x - y subtracts y from x in pretty much any language that supports operators.

      • Yggstyle@lemmy.world
        4·
        7 days ago

        Technically yes… But I think he was more making the excuse for the gore “from the goresmith’s perspective.”

        And I’m not sure if the compiler in any language would change a random check function… The others are a possibility.

      • Hirom@beehaw.org
        2·
        6 days ago

        An infinite loop canot be ruled out in the last case, so a compiler couldn’t optimize this away without potentially changing the program behavior.

            • yetAnotherUser@discuss.tchncs.de
              1·
              6 days ago

              Even though this isn’t C, but if we take from the C11 draft §6.8.5 point 6 (https://www.open-std.org/jtc1/sc22/wg14/www/docs/n1570.pdf):

              An iteration statement whose controlling expression is not a constant expression, that performs no input/output operations, does not access volatile objects, and performs no synchronization or atomic operations in its body, controlling expression, or (in the case of a for statement) its expression-3, may be assumed by the implementation to terminate

              “new Random().nextInt()” might perform I/O though so it could still be defined behavior. Or the compiler does not assume this assumption.

              But an aggressive compiler could realize the loop would not terminate if x does not become 10 so x must be 10 because the loop can be assumed to terminate.