使用syn时如何从Option <T>获取T？

Question

我syn用来解析Rust代码。当我使用读取命名字段的类型时field.ty，得到一个syn::Type。当我使用打印时，quote!{#ty}.to_string()我得到了"Option<String>"。

我怎样才能得到"String"？我想用#ty在quote!打印"String"代替"Option<String>"。

我想生成如下代码：

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

从...开始

struct Foo {
    bar: Option<String>
}

我的尝试：

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

我对来自 @Boiethios 的响应的更新版本，在公共板条箱中进行了测试和使用，支持以下几种语法Option：

• Option
• std::option::Option
• ::std::option::Option
• core::option::Option
• ::core::option::Option

fn extract_type_from_option(ty: &syn::Type) -> Option<&syn::Type> {
    use syn::{GenericArgument, Path, PathArguments, PathSegment};

    fn extract_type_path(ty: &syn::Type) -> Option<&Path> {
        match *ty {
            syn::Type::Path(ref typepath) if typepath.qself.is_none() => Some(&typepath.path),
            _ => None,
        }
    }

    // TODO store (with lazy static) the vec of string
    // TODO maybe optimization, reverse the order of segments
    fn extract_option_segment(path: &Path) -> Option<&PathSegment> {
        let idents_of_path = path
            .segments
            .iter()
            .into_iter()
            .fold(String::new(), |mut acc, v| {
                acc.push_str(&v.ident.to_string());
                acc.push('|');
                acc
            });
        vec!["Option|", "std|option|Option|", "core|option|Option|"]
            .into_iter()
            .find(|s| &idents_of_path == *s)
            .and_then(|_| path.segments.last())
    }

    extract_type_path(ty)
        .and_then(|path| extract_option_segment(path))
        .and_then(|path_seg| {
            let type_params = &path_seg.arguments;
            // It should have only on angle-bracketed param ("<String>"):
            match *type_params {
                PathArguments::AngleBracketed(ref params) => params.args.first(),
                _ => None,
            }
        })
        .and_then(|generic_arg| match *generic_arg {
            GenericArgument::Type(ref ty) => Some(ty),
            _ => None,
        })
}

Answer 2

您应该执行以下未经测试的示例：

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"),
    }
}

没有太多要解释的东西，它只是“展开”类型的各种组成部分：

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

如果有更简单的方法可以做到，那么我将很高兴知道。

请注意，由于syn是解析器，因此可以使用令牌。您无法确定这是不是Option。用户可以例如输入std::option::Option或type MaybeString = std::option::Option<String>;。您不能处理这些任意名称。