Fix namespace name conflict detection in "Convert named imports to namespace import" action #45019
Conversation
typescript-bot
added
the
For Milestone Bug
| if (!neededNamedImports.some(n => n.name === element.name)) { | ||
| neededNamedImports.push(factory.createImportSpecifier(element.propertyName && factory.createIdentifier(element.propertyName.text), factory.createIdentifier(element.name.text))); | ||
| } | ||
| else if (isExportSpecifier(id.parent)) { |
There was a problem hiding this comment.
Old code disregarded named export clauses that aliased the reference to the named import, so code such as
import { a } from "m";
export { a as b };was in fact incorrectly refactored into
import { a } from "m";
export { m.a as b };
| const neededNamedImports: ImportSpecifier[] = []; | ||
| // Imports that need to be kept as named imports in the refactored code, to avoid changing the semantics. | ||
| // More specifically, those are named imports that appear in named exports in the original code, e.g. `a` in `import { a } from "m"; export { a }`. | ||
| const neededNamedImports: Set<ImportSpecifier> = new Set(); |
There was a problem hiding this comment.
Changed to use a set instead of an array. The old code used the array as a set by comparing the named imports' name properties (of type Identifiers) (element.name in for loop below), but it pushed into the array a node with a newly-created Identifier as name, so in fact this new node's name would never be equal to the named imports' name, and therefore the same named import could be repeated, as in:
import { a, b } from "m";
export { b };
export { b as c };would become
import * as m from "m";
import { b, b } from "m"; // Duplicate b import
export { b };
export { b as c };
gabritto
changed the title
andrewbranch
left a comment
andrewbranch
left a comment
There was a problem hiding this comment.
🌟 Thanks for the great code comments! Made it really easy to understand.
Fixes #44267