slim estuary
For example if I have an existing type
Foo<'bar, B: Baz, W: Wut = ()> { ... }
and I have a struct I want to create
struct MyType<?> {
foo: Foo<?>,
}
What can I do to scrape in all the ? and just use it verbatim in Foo? Is that a thing?
uneven depot
You can't. A type's generic list can't be dependent on another type's list; you need to specify them explicitly
slim estuary
I can't quite figure out which direction the "dependant" goes in your answer. Do you mean that the ? in Foo<?> must be explicit? Like Foo<ConcreteType>?
uneven depot
No. You can't make MyType's generic list inherit Foo's generic list, or the other way around: they have to be declared separately
You have to manually do this: ```rust
struct MyType<'bar, B: Baz, W: Wut = ()> {
foo: Foo<'bar, B, W>,
}
ooh. ok! that makes sense. it would make no sense for the W: Wut = () to appear inside the struct. I guess it's analogous to a function's parameters (in the declaration) vs arguments (in the body).
tender rock
depending on what you're doing, it might be possible to work with a
struct MyType<F> {
foo: F,
}
and then you only need to mention the Foo type and its parameters where they are relevant (such as in impl blocks), rather than in every single use of MyType
uneven depot
That too ^