fixes #19: provides fallback T for ExtractValues<T>

Author: everweij

package.json

@@ -1,7 +1,7 @@
 {
   "private": false,
   "name": "typescript-result",
-  "version": "3.5.0",
+  "version": "3.5.1",
   "description": "Supercharge your TypeScript error handling with a powerful Result type that transforms chaotic try-catch blocks into elegant, type-safe code.",
   "keywords": [
     "result",

src/result.test.ts

@@ -2862,6 +2862,23 @@ describe("Result", () => {
 		);
 		expectTypeOf(resultE).toEqualTypeOf<AsyncResult<number, never>>();
 	});
+
+	it("Correctly infers the ok-value when there's some overlap with an Result instance", () => {
+		// See: https://github.com/everweij/typescript-result/issues/19
+
+		// { value: string } has some overlap with the Result type, so it seems like
+		// TS is very strict with `ReturningValue<T>` and returns never. The workaround seem
+		// to be to give `ExtractValue<T>` a fallback of `T`.
+		const resultA = Result.ok(12).map(() => ({ value: "bar" }));
+
+		expectTypeOf(resultA).toEqualTypeOf<Result.Ok<{ value: string }>>();
+
+		const resultB = Result.try(() => ({ data: { value: "string" } })).map(
+			({ data }) => data,
+		);
+
+		expectTypeOf(resultB).toEqualTypeOf<Result<{ value: string }, Error>>();
+	});
 });
 
 describe("AsyncResult", () => {

src/result.ts

@@ -42,7 +42,7 @@ type ReturningError<T> =
 	| AsyncResult<any, T>
 	| Promise<ReturningError<T>>;
 
-type ExtractValue<T> = T extends ReturningValue<infer Value> ? Value : never;
+type ExtractValue<T> = T extends ReturningValue<infer Value> ? Value : T;
 type ExtractError<T> = T extends ReturningError<infer Error> ? Error : never;
 
 type ExtractValues<T extends any[]> = {