3 ms·
There is in c++, too (std::ignore). Not sure why the author decided to go with the ancient void cast
by on_the_train 10mo ago
There is in c++, too (std::ignore). Not sure why the author decided to go with the ancient void cast
- aw1621107 10mo agoI believe C++26 now allows _ as a placeholder name [0]: > We propose that when_ is used as the identifier for the declaration of a variable, non static class member variable, lambda capture or structured binding. the introduced name is implicitly given the [[maybe_unused]] attribute. > In contexts where the grammar expects a pattern matching pattern,_ represents the wildcard pattern. Some of the finer details (e.g., effect on lifetime and whether a _ variable can be used) differ, though. [0]: https://github.com/cplusplus/papers/issues/878 https://github.com/cplusplus/papers/issues/878
- tialaramex 10mo agoSpecifically, Rust's _ is not a variable, it is a pattern that matches anything and so let _ = isn't an assignment it's specifically the explicit choice not to assign this value. If we wrote a "dummy" variable the compiler is forbidden from dropping the value early, that "dummy" is alive until it leaves scope, whereas if we never assigned the value it's dropped immediately. In modern Rust you don't need the let here because you're allowed to do the pattern match anywhere, and as I said _ is simply a pattern that matches anything. So we could omit the let keyword, but people don't.
- vitus 10mo agostd::ignore's behavior outside of use with std::tie is not specified in any finalized standard. https://www.open-std.org/jtc1/sc22/wg21/docs/papers/2023/p2968r2.html https://www.open-std.org/jtc1/sc22/wg21/docs/papers/2023/p29... aims to address that, but that won't be included until C++26 (which also includes _ as a sibling commenter mentions).
- quuxplusone 10mo agohttps://quuxplusone.github.io/blog/2022/10/16/prefer-core-over-library/ https://quuxplusone.github.io/blog/2022/10/16/prefer-core-ov... does not (yet) include this as one of its examples, but it could (and someday might).