Reject ISO week 53 in years that only have 52 weeks

arrow/util.py

@@ -101,6 +101,12 @@ def iso_to_gregorian(iso_year: int, iso_week: int, iso_day: int) -> datetime.dat
     year_start = fourth_jan - delta
     gregorian = year_start + datetime.timedelta(days=iso_day - 1, weeks=iso_week - 1)
 
+    # Only some years have 53 ISO weeks. For the ones that don't, the
+    # arithmetic above quietly lands in the next ISO year rather than failing,
+    # so reject a week the requested year does not actually have.
+    if gregorian.isocalendar()[0] != iso_year:
+        raise ValueError(f"ISO Calendar year {iso_year} has no week {iso_week}.")
+
     return gregorian
 
 

tests/test_factory.py

@@ -236,6 +236,11 @@ def test_one_arg_iso_calendar(self):
         with pytest.raises(ValueError):
             self.factory.get((2014, 7, 10))
 
+        # 2025 has 52 ISO weeks, so week 53 is not a date in that year and
+        # must not silently become the first week of 2026
+        with pytest.raises(ValueError):
+            self.factory.get((2025, 53, 1))
+
     def test_one_arg_other(self):
         with pytest.raises(TypeError):
             self.factory.get(object())

tests/test_util.py

@@ -131,3 +131,15 @@ def test_iso_gregorian(self):
 
         with pytest.raises(ValueError):
             util.iso_to_gregorian(2013, 8, 0)
+
+    def test_iso_gregorian_week_53(self):
+        # 2026 is one of the years that has 53 ISO weeks
+        assert util.iso_to_gregorian(2026, 53, 1) == datetime(2026, 12, 28).date()
+
+        # 2025 only has 52, so week 53 must be rejected rather than rolling
+        # over into the first week of 2026
+        with pytest.raises(ValueError):
+            util.iso_to_gregorian(2025, 53, 1)
+
+        with pytest.raises(ValueError):
+            util.iso_to_gregorian(2024, 53, 1)