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Can you help me understand how references work when calling methods? (feddit.it)

submitted 5 days ago by dontblink@feddit.it to c/rust@lemmy.ml

4 comments fedilink hide all child comments

This is what I've found:

In C and C++, two different operators are used for calling methods: you use . if you’re calling a method on the object directly and -> if you’re calling the method on a pointer to the object and need to dereference the pointer first. In other words, if object is a pointer, object->something() is similar to (*object).something().

What about Rust?

Rust doesn’t have an equivalent to the -> operator. Instead, Rust has what is called automatic referencing and dereferencing.

In other words, the following are the same:

p1.distance(&p2); (&p1).distance(&p2); Note: this automatic referencing behavior works because methods have a clear receiver-the type of self. Given the receiver and name of a method, Rust can figure out definitively whether the method is reading (&self), mutating (&mut self), or consuming (self).

I am not sureI understood the note correctly, what does it mean that there is a clear receiver? Also, doesn't Rust actually have an operator for dereferencing (*) as well？

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[–] nous@programming.dev 4 points 5 days ago (2 children)

By clear receiver it means there is only one function a name can point to. For instance you cannot have:

struct Foo;
impl Foo {
    pub fn foo(self) {}
    pub fn foo(&amp;self) {}
}

error[E0592]: duplicate definitions with name `foo`
 --> src/lib.rs:5:5
  |
4 |     pub fn foo(self) {}
  |     ---------------- other definition for `foo`
5 |     pub fn foo(&amp;self) {}
  |     ^^^^^^^^^^^^^^^^^ duplicate definitions for `foo`

Which means it is easy to see what the function requires when you call it. It will either have self or &self and when it is &self it can auto reference the value for you. This is unlike C and C++ where you can overload a function definition with multiple different signatures and bodies and thus the signature is important to know to know which function to actually call.

Yes rust has a operator for dereferencing as well (the *). This can be used to copy a value out of a reference for simple types the implement Copy at least.

[–] TehPers 2 points 5 days ago

Everything you said makes sense, but just want to add to your note at the bottom: * can also be used for reborrowing (&*blah, &mut *blah, &***blah, etc). This is useful for getting a shared borrow (&T) or unique/exclusive borrow (&mut T) from another type of pointer (like Box, Arc, or even MutexGuard).

[–] Ephera@lemmy.ml 2 points 5 days ago (1 children)

And if you're wondering, what the hell, if there's no overloading, how do you even design an API???
You'll learn about the From and Into traits, which allow you to take multiple types for a parameter, because those traits implement the conversion between those types.

[–] beeb@lemm.ee 2 points 5 days ago* (last edited 5 days ago)

If you want polymorphism which looks more like what you're describing, you can put trait bounds on parameters instead of a type and it will accept any parameter that implement those traits. E.g. If you want to accept anything that can be turned into an owned string with ".into()" you type an argument with "impl Into". Another common one is "impl AsRef" to accept a path, path reference, PathBuf etc.