• DickFiasco@sh.itjust.works
    7·
    21 days ago

    I see. I think auto gets overused a lot by people just being lazy about writing out the full type, but constexpr is good practice in my opinion. Never had a compiler issue with them, but then I don’t think I’ve ever used a nested lambda either.

    • thebestaquaman@lemmy.world
      2·
      21 days ago

      I really like C++ (I know, shoot me), and I think auto should be avoided at (almost) all costs.

      One of the things I love about a language like C++ is that I can take one glance at the code and immediately know what types I’m working with. auto takes that away while adding almost no benefit outside of a little convenience while writing.

      If I’m working with some very big template type that I don’t want to write out, 99/100 times I’ll just have a using somewhere to make it more concise. Hell, I’ll have using vectord = std::vector<double> if I’m using a lot of them, because I think it makes the code more readable. Just don’t throw auto at me.

      Of course, the worst thing ever (which I’ve seen far too often) is the use of auto in examples in documentation. Fucking hell! I’m reading the docs because I don’t know the library well! When you first bother to write examples, at least let me know the return type without needing to dig through your source code!

        • thebestaquaman@lemmy.world
          1·
          21 days ago

          Thanks, that was a good read :)

          However, my impression is that he’s largely using the existence of templates and polymorphism as arguments that “we don’t really care about type”. I disagree: A template is essentially a generic type description that says something about what types are acceptable. When working with something polymorphic, I’ll prefer ParentClass&, to indicate what kind of interface I’m working with.

          Sure, it can be very useful to hide exact type information in order to generalise the code, but I think that’s a weak argument for hiding all type information by default, which is what auto does.

            • thebestaquaman@lemmy.world
              0·
              19 days ago

              Similarly, what would you gain by saying uint32_t const* x = my_var.get<uint32_t>();

              To be frank: You gain the information that MyConcreteType::get<uint32_t> returns a uint32_t, which I otherwise couldn’t infer from the docs. Of course, I could assume it, based on the template parameter, but I don’t want to go around assuming a bunch of stuff in order to read docs.

              Take an example like auto x = my_var.to_reduced_form(), it’s very clear that x is the “reduced form” of my_var, which could be meaningful in itself, but what type is it? I need to know that if I want to do anything with x. Can I do x += 1? If I do, will that modify my_var? Let’s say I want to make a vector of whatever to_reduced_form returns… and so on.

              All these questions are very easily answered by MyConcreteType x = my_var.to_reduced_form(). Now I immediately know that everything I can do with my_var, I can also do with x. This makes me happy, because I need to do less digging, and the code becomes clearer to read.

    • voodooattack@lemmy.worldOP
      0·
      20 days ago

      The problem is that lambdas with a capture aren’t strongly typed are uniquely typed, so you have to use decltype/auto. And if you pass such a lambda to a function you’ll have to use auto as well.

      If you write a lambda with a capture that calls itself recursively you’ll have to pass it to itself as an auto argument as part of the call signature.

      I think this article explains it better: https://artificial-mind.net/blog/2020/09/12/recursive-lambdas

      Edit: fixed wrong terminology