• @[email protected]
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    944 days ago

    Except that many other languages have proven that C++ is simply terrible at providing meaningful errors.

    • @[email protected]
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      3 days ago

      The whole point of a segfault is that you can’t really know anything about it. Even in rust, when you get a segfault there is no meaningful error.

      • @[email protected]
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        73 days ago

        point is, Rust manages to give you not a segfault but a meaningful error almost all the time until you use unsafe

        • @[email protected]
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          -23 days ago

          If you’re getting a segfault in C++, it’s also cause you used unsafe code. It’s just not officially enclosed in an “unsafe” block.

          • @[email protected]
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            2 days ago

            the point was not on the unsafe word, but a very specific feature of Rust that helps enclosing unsafe code where the compiler wouldn’t be able to 100% verify your logic. no such thing in C++. C++ does not even attempt to verify safety

            your response is basically “get better at coding dumbass, I am the safety validator”

      • @[email protected]
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        42 days ago

        The worst thing you can do in non-unsafe Rust is perform an out-of-bounds indexing operation, or anything else that panics. The error you get tells you the panic’s exact location in the source code, down to the line and column. Meanwhile, C and C++ either don’t produce an error at all when accessing uninitialized memory (which is arguably the worst behavior), or it segfaults with zero extra info.

        The only way to make Rust segfault is by performing unsafe operations, and those must always be clearly marked.

        • @[email protected]
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          2 days ago

          The only way to make Rust segfault is by performing unsafe operations.

          Challange accepted. The following Rust code technically segfaults:

          fn stackover(a : i64) -> i64 {
              return stackover(a);
          }
          
          
          fn main() {
              println!("{}", stackover(100));
          }
          

          A stack overflow is technically a segmentation violation. At least on linux the program recives the SIGSEGV signal. This compiles and I am no rust dev but this does not use unsafe code, right?

          While the compiler shows a warning, the error message the program prints when run is not very helpfull IMHO:

          thread 'main' has overflowed its stack
          fatal runtime error: stack overflow
          [1]    45211 IOT instruction (core dumped)  ../target/debug/rust
          

          Edit: Even the compiler warning can be tricked by making it do recusion in pairs:

          fn stackover_a(a : i64) -> i64 {
              return stackover_b(a);
          }
          
          fn stackover_b(a : i64) -> i64 {
              return stackover_a(a);
          }
          
          fn main() {
              println!("{}", stackover_a(100));
          }