Fix subsetEquality: same referenced object on same level node of tree is regarded as circular reference

Author: Gianfranco

packages/expect/src/__tests__/utils.test.js

@@ -333,6 +333,19 @@ describe('subsetEquality()', () => {
       expect(subsetEquality(primitiveInsteadOfRef, circularObjA1)).toBe(false);
     });
 
+    test('referenced object on same level should not regarded as circular reference', () => {
+      const referencedObj = {abc: 'def'};
+      const object = {
+        a: {abc: 'def'},
+        b: {abc: 'def', zzz: 'zzz'},
+      };
+      const thisIsNotCircular = {
+        a: referencedObj,
+        b: referencedObj,
+      };
+      expect(subsetEquality(object, thisIsNotCircular)).toBeTruthy();
+    });
+
     test('transitive circular references', () => {
       const transitiveCircularObjA1 = {a: 'hello'};
       transitiveCircularObjA1.nestedObj = {parentObj: transitiveCircularObjA1};

packages/expect/src/utils.ts

@@ -166,7 +166,6 @@ export const iterableEquality = (
   if (a.constructor !== b.constructor) {
     return false;
   }
-
   let length = aStack.length;
   while (length--) {
     // Linear search. Performance is inversely proportional to the number of
@@ -290,20 +289,25 @@ export const subsetEquality = (
 
     return Object.keys(subset).every(key => {
       if (isObjectWithKeys(subset[key])) {
-        if (seenReferences.get(subset[key])) {
+        if (seenReferences.has(subset[key])) {
           return equals(object[key], subset[key], [iterableEquality]);
         }
         seenReferences.set(subset[key], true);
       }
-
-      return (
+      const result =
         object != null &&
         hasOwnProperty(object, key) &&
         equals(object[key], subset[key], [
           iterableEquality,
           subsetEqualityWithContext(seenReferences),
-        ])
-      );
+        ]);
+      // The main goal of using seenReference is to avoid circular node on tree.
+      // It will only happen within a parent and its child, not a node and nodes next to it (same level)
+      // We should keep the reference for a parent and its child only
+      // Thus we should delete the reference immediately so that it doesn't interfere
+      // other nodes within the same level on tree.
+      seenReferences.delete(subset[key]);
+      return result;
     });
   };