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[ty] Make Module::all_submodules return Module instead of Name

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This is to facilitate recursive traversal of all modules in an
environment. This way, we can keep asking for submodules.

This also simplifies how this is used in completions, and probably makes
it faster. Namely, since we return the `Module` itself, callers don't
need to invoke the full module resolver just to get the module type.

Note that this doesn't include namespace packages. (Which were
previously not supported in `Module::all_submodules`.) Given how they
can be spread out across multiple search paths, they will likely require
special consideration here.
This commit is contained in:
Andrew Gallant 2025-08-28 10:53:53 -04:00 • committed by Andrew Gallant
parent 9cea752934
commit 046893c186
3 changed files with 74 additions and 37 deletions
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15
crates/ty_python_semantic/src/semantic_model.rs
View file

@ -173,19 +173,10 @@ impl<'db> SemanticModel<'db> {
let builtin = module.is_known(self.db, KnownModule::Builtins);
let mut completions = vec![];
for submodule_basename in module.all_submodules(self.db) {
let Some(basename) = ModuleName::new(submodule_basename.as_str()) else {
continue;
};
let mut submodule_name = module.name(self.db).clone();
submodule_name.extend(&basename);
let Some(submodule) = resolve_module(self.db, &submodule_name) else {
continue;
};
let ty = Type::module_literal(self.db, self.file, submodule);
for submodule in module.all_submodules(self.db) {
let ty = Type::module_literal(self.db, self.file, *submodule);
completions.push(Completion {
name: submodule_basename.clone(),
name: Name::new(submodule.name(self.db).as_str()),
ty,
builtin,
});