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[ty] Make Module a Salsa ingredient

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We want to write queries that depend on `Module` for caching. While it
seems it can be done without making `Module` an ingredient, it seems it
is best practice to do so.

[best practice to do so]: https://github.com/astral-sh/ruff/pull/19408#discussion_r2215867301
This commit is contained in:
Andrew Gallant 2025-07-18 14:15:42 -04:00 • committed by Andrew Gallant
parent 905b9d7f51
commit 4573a0f6a0
24 changed files with 258 additions and 253 deletions
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6
crates/ty_python_semantic/src/semantic_model.rs
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@ -67,8 +67,8 @@ impl<'db> SemanticModel<'db> {
tracing::debug!("Could not resolve module from `{module_name:?}`");
return vec![];
};
let ty = Type::module_literal(self.db, self.file, &module);
let builtin = module.is_known(KnownModule::Builtins);
let ty = Type::module_literal(self.db, self.file, module);
let builtin = module.is_known(self.db, KnownModule::Builtins);
let mut completions = vec![];
for crate::types::Member { name, ty } in crate::types::all_members(self.db, ty) {
@ -84,7 +84,7 @@ impl<'db> SemanticModel<'db> {
let Some(submodule) = resolve_module(self.db, &submodule_name) else {
continue;
};
let ty = Type::module_literal(self.db, self.file, &submodule);
let ty = Type::module_literal(self.db, self.file, submodule);
completions.push(Completion {
name: submodule_basename.clone(),
ty,