Does Rust erase generic types or not?

Question

Is there type erasure of generics in Rust (like in Java) or not? I am unable to find a definitive answer.

Answer 1

When you use a generic function or a generic type, the compiler generates a separate instance for each distinct set of type parameters (I believe lifetime parameters are ignored, as they have no influence on the generated code). This process is called monomorphization. For instance, Vec<i32> and Vec<String> are different types, and therefore Vec<i32>::len() and Vec<String>::len() are different functions. This is necessary, because Vec<i32> and Vec<String> have different memory layouts, and thus need different machine code! Therefore, no, there is no type erasure.

If we use Any::type_id(), as in the following example:

use std::any::Any;

fn main() {
    let v1: Vec<i32> = Vec::new();
    let v2: Vec<String> = Vec::new();
    
    let a1 = &v1 as &dyn Any;
    let a2 = &v2 as &dyn Any;
    
    println!("{:?}", a1.type_id());
    println!("{:?}", a2.type_id());
}

we obtain different type IDs for two instances of Vec. This supports the fact that Vec<i32> and Vec<String> are distinct types.

However, reflection capabilities in Rust are limited; Any is pretty much all we've got for now. You cannot obtain more information about the type of a runtime value, such as its name or its members. In order to be able to work with Any, you must cast it (using Any::downcast_ref() or Any::downcast_mut() to a type that is known at compile time.