tsparser: usage of default exported resources #1970
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All committers have signed the CLA. |
fredr
changed the title
eandre
reviewed
Jun 18, 2025
eandre
reviewed
Jun 18, 2025
eandre
reviewed
Jun 18, 2025
| @@ -39,7 +39,7 @@ pub const GATEWAY_PARSER: ResourceParser = ResourceParser { | |||
| for r in iter_references::<Res>(&module, &names) { | |||
| let r = report_and_continue!(r); | |||
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| let object = match &r.bind_name { | |||
| let object = match r.bind_name.ident() { | |||
| None => None, | |||
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I guess this could resolve the default export, but not sure how difficult that would be to do. If it's tricky maybe add a TODO here?
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I'm not entirely sure how object is used, but I've given it a stab, it resolves to something, not sure exactly what we want it to resolve to?
| @@ -61,7 +61,7 @@ fn parse_cron_job( | |||
| pass: &mut ResourceParseContext, | |||
| r: NamedClassResource<DecodedCronJobConfig>, | |||
| ) -> ParseResult<()> { | |||
| let object = match &r.bind_name { | |||
| let object = match r.bind_name.ident() { | |||
| @@ -46,7 +46,7 @@ pub const OBJECTS_PARSER: ResourceParser = ResourceParser { | |||
| let r = report_and_continue!(r); | |||
| let cfg = r.config.unwrap_or_default(); | |||
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| let object = match &r.bind_name { | |||
| let object = match r.bind_name.ident() { | |||
| }); | ||
| } | ||
| } | ||
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| { | ||
| for r in iter_references::<NamedStaticMethod>(&module, &names) { | ||
| let r = report_and_continue!(r); | ||
| let object = match &r.bind_name { | ||
| let object = match r.bind_name.ident() { |
| @@ -71,7 +71,7 @@ pub const SUBSCRIPTION_PARSER: ResourceParser = ResourceParser { | |||
| spread.err("cannot use ... for PubSub topic reference"); | |||
| continue; | |||
| } | |||
| let object = match &r.bind_name { | |||
| let object = match r.bind_name.ident() { | |||
| @@ -63,7 +63,7 @@ pub const TOPIC_PARSER: ResourceParser = ResourceParser { | |||
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| for r in iter_references::<PubSubTopicDefinition>(&module, &names) { | |||
| let r = report_and_continue!(r); | |||
| let object = match &r.bind_name { | |||
| let object = match r.bind_name.ident() { | |||
| @@ -152,7 +152,7 @@ pub const SQLDB_PARSER: ResourceParser = ResourceParser { | |||
| } | |||
| }; | |||
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| let object = match &r.bind_name { | |||
| let object = match r.bind_name.ident() { | |||
| }); | ||
| } | ||
| } | ||
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| { | ||
| for r in iter_references::<NamedStaticMethod>(&module, &names) { | ||
| let r = report_and_continue!(r); | ||
| let object = match &r.bind_name { | ||
| let object = match r.bind_name.ident() { |
eandre
reviewed
Jun 18, 2025
| Some(id) => pass | ||
| let object = match r.bind_name { | ||
| BindName::Anonymous => None, | ||
| BindName::DefaultExport(ref expr) => { |
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I don't think we should be resolving this by passing in an expression to resolve_obj.
resolve_obj goes into self.expr(...) which goes into a big match statement where we end up at:
ast::Expr::Ident(expr) => {
let Some(obj) = self.ident_obj(expr) else {
HANDLER.with(|handler| handler.span_err(expr.span, "unknown identifier"));
return Type::Basic(Basic::Never);
};
let named = Named::new(obj, vec![]);
Type::Named(named)
}This goes into ident_obj(expr):
fn ident_obj(&self, ident: &ast::Ident) -> Option<Rc<Object>> {
// Does this represent a type parameter?
self.state.resolve_module_ident(self.module, ident)
}
}And resolve_module_ident checks:
// Is it a top-level object in this module?
let ast_id = AstId::from(ident);
if let Some(obj) = module.data.top_level.get(&ast_id) {
return Some(obj.clone());
}I believe the correct way to resolve default exports is to do the same thing, but check:
if let Some(obj) = module.data.default_export {
return Some(obj.clone());
}(sorry for the trouble)
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the resolved bind of a default export will have no name, but we still need to add it or it wont be tracked