impl<T> seems to be redundant

1

I believe that if we have:

struct AStruct<T>
{
}

we should be able to do:
impl AStruct<T>//here notice no impl<T> as it is simply in my opinion redundant
{
}

Any thoughts?

2

In that case it would not be immediately obvious if T is an existing type or type parameter.

3

I believe that it is immediately obvious by having:

AStruct<T>
4 
struct T;

struct AStruct<T> {}

impl AStruct<T> {
    fn foo() {}
}

would only implement the foo method for the type AStruct<T> where T is a struct. It wouldn't implement it for AStruct<T> for all types T like impl<T> AStruct<T> does.

5

Is this valid rust?

6
7

The impl<T> part declares a generic type variables and the AStruct<T> part uses it. This separation allows more complex impl blocks:

impl<T> AStruct<Option<T>> {}

impl<T, E> AStruct<Result<T, E>> {}

impl<T> AStruct<Box<dyn Iterator<Item = T>>> {}

impl AStruct<String> {}

...
8

Thanks, so do I understand it correctly that having the below:

struct T;
impl aStruct<T>
{
}

aStruct type will work only for struct T? Some sort of specialization?

9

Conceptually, yes. But usually when we refer the term specialization it means this WIP unstable feature.