How can I get the T from an Option<T> when using syn?

Question

I'm using syn to parse Rust code. When I read a named field's type using field.ty, I get a syn::Type. When I print it using quote!{#ty}.to_string() I get "Option<String>".

How can I get just "String"? I want to use #ty in quote! to print "String" instead of "Option<String>".

I want to generate code like:

impl Foo {
    pub set_bar(&mut self, v: String) {
        self.bar = Some(v);
    }
}

starting from

struct Foo {
    bar: Option<String>
}

My attempt:

let ast: DeriveInput = parse_macro_input!(input as DeriveInput);

let data: Data = ast.data;

match data {
    Data::Struct(ref data) => match data.fields {
        Fields::Named(ref fields) => {

            fields.named.iter().for_each(|field| {
                let name = &field.ident.clone().unwrap();

                let ty = &field.ty;
                quote!{
                    impl Foo {
                        pub set_bar(&mut self, v: #ty) {
                            self.bar = Some(v);
                        }
                    }
                };      
            });
        }
        _ => {}
    },
    _ => panic!("You can derive it only from struct"),
}

Answer 1

You should do something like this untested example:

use syn::{GenericArgument, PathArguments, Type};

fn extract_type_from_option(ty: &Type) -> Type {
    fn path_is_option(path: &Path) -> bool {
        leading_colon.is_none()
            && path.segments.len() == 1
            && path.segments.iter().next().unwrap().ident == "Option"
    }

    match ty {
        Type::Path(typepath) if typepath.qself.is_none() && path_is_option(typepath.path) => {
            // Get the first segment of the path (there is only one, in fact: "Option"):
            let type_params = typepath.path.segments.iter().first().unwrap().arguments;
            // It should have only on angle-bracketed param ("<String>"):
            let generic_arg = match type_params {
                PathArguments::AngleBracketed(params) => params.args.iter().first().unwrap(),
                _ => panic!("TODO: error handling"),
            };
            // This argument must be a type:
            match generic_arg {
                GenericArgument::Type(ty) => ty,
                _ => panic!("TODO: error handling"),
            }
        }
        _ => panic!("TODO: error handling"),
    }
}

There's not many things to explain, it just "unrolls" the diverse components of a type:

Type -> TypePath -> Path -> PathSegment -> PathArguments -> AngleBracketedGenericArguments -> GenericArgument -> Type.

If there is an easier way to do that, I would be happy to know it.

---

Note that since syn is a parser, it works with tokens. You cannot know for sure that this is an Option. The user could, for example, type std::option::Option, or write type MaybeString = std::option::Option<String>;. You cannot handle those arbitrary names.