Function filterArray

assert/object_match.ts:159–223

(a: unknown[], b: unknown[])

  }

  function filterArray(a: unknown[], b: unknown[]): unknown[] {
    // Prevent infinite loop with circular references with same filter
    const memo = seen.get(a);
    if (memo && (memo === b)) return a;

    seen.set(a, b);

    const filtered: unknown[] = [];
    const count = Math.min(a.length, b.length);

    for (let i = 0; i < count; ++i) {
      const value = a[i];
      const subset = b[i];

      // On regexp references, keep value as it to avoid losing pattern and flags
      if (value instanceof RegExp) {
        filtered.push(value);
        continue;
      }
      // On date references, keep value as it to avoid losing the timestamp
      if (value instanceof Date) {
        filtered.push(value);
        continue;
      }

      // On array references, build a filtered array and filter nested objects inside
      if (Array.isArray(value) && Array.isArray(subset)) {
        filtered.push(filterArray(value, subset));
        continue;
      }

      // On nested objects references, build a filtered object recursively
      if (isObject(value) && isObject(subset)) {
        // When both operands are maps, build a filtered map with common keys and filter nested objects inside
        if ((value instanceof Map) && (subset instanceof Map)) {
          const map = new Map(
            [...value].filter(([k]) => subset.has(k))
              .map(([k, v]) => {
                const v2 = subset.get(k);
                if (isObject(v) && isObject(v2)) {
                  return [k, filterObject(v as Loose, v2 as Loose)];
                }

                return [k, v];
              }),
          );
          filtered.push(map);
          continue;
        }

        // When both operands are set, build a filtered set with common values
        if ((value instanceof Set) && (subset instanceof Set)) {
          filtered.push(value.intersection(subset));
          continue;
        }

        filtered.push(filterObject(value as Loose, subset as Loose));
        continue;