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You mean there is no compile time way to accomplish that? Doesn't sound very Rust-like.
by tus88 7y ago
You mean there is no compile time way to accomplish that? Doesn't sound very Rust-like.
- jeffadotio 7y ago“Option” is the operative word. A type that does not implement the correct trait cannot be used in a way that requires the trait. If you are writing anything non-trivial that would probably be preferable. It requires nothing, that is the default behavior, so I did not bother to mention it.
- re 7y agoIt is determined at compile time; I'm not sure why GP is suggesting returning false if you don't want types to be compared. By default structs are not comparable. If you opt-in to the standard PartialEq implementation using the `derive` annotation[1], it will only allow a struct to be compared for equality with other structs of the same type. You can choose to add implementations to compare against other types[2]. [1] https://doc.rust-lang.org/book/ch05-02-example-structs.html#adding-useful-functionality-with-derived-traits https://doc.rust-lang.org/book/ch05-02-example-structs.html#... [2] https://doc.rust-lang.org/std/cmp/trait.PartialEq.html#how-can-i-compare-two-different-types https://doc.rust-lang.org/std/cmp/trait.PartialEq.html#how-c...
- artursapek 7y agoRust is such a delightful language.
- aliceryhl 7y agoI don't know why they would make it always return false. In Rust you can just not implement Eq for the type, and it becomes a compile error to try and compare it.