Fix TimeWithinDay(t) abstract operation (#3974)

Fixes #3973.

TimeWithinDay(t) = t modulo msPerDay, where msPerDay = 86400000.
The sign of the modulo operation result should be same as the right side
per definition, conseqently TimeWithinDay(t) >= 0.

References:
- https://www.ecma-international.org/ecma-262/11.0/#sec-mathematical-operations
- https://www.ecma-international.org/ecma-262/11.0/#sec-overview-of-date-objects-and-definitions-of-abstract-operations

JerryScript-DCO-1.0-Signed-off-by: Csaba Osztrogonác oszi@inf.u-szeged.hu

jerry-core/ecma/builtin-objects/ecma-builtin-helpers-date.c

@@ -72,7 +72,13 @@ ecma_date_time_within_day (ecma_number_t time) /**< time value */
 {
   JERRY_ASSERT (!ecma_number_is_nan (time));
 
-  return (ecma_number_t) fmod (time, ECMA_DATE_MS_PER_DAY);
+  ecma_number_t modulo = fmod (time, ECMA_DATE_MS_PER_DAY);
+  if (modulo < 0)
+  {
+    modulo += ECMA_DATE_MS_PER_DAY;
+  }
+
+  return modulo;
 } /* ecma_date_time_within_day */
 
 /**

tests/jerry/date-setters.js

@@ -207,6 +207,23 @@ d.setTime(0);
 assert (d.setUTCFullYear(1970) == 0);
 assert (d.getUTCFullYear() == 1970);
 
+/* ECMA262 v11 20.4.1.2 Day Number and Time within Day
+   msPerDay = 86400000
+   TimeWithinDay(t) = t modulo msPerDay
+
+   ECMA262 v11 5.2.5 Mathematical Operations
+   The notation “x modulo y” (y must be finite and nonzero) computes a value k of the same sign as y (or zero).
+
+   Consequently TimeWithinDay(t) >= 0. It can be tested properly with dates close to 1970.
+*/
+d = new Date("1969-12-01T01:00:00.000Z");
+d.setFullYear(1968);
+assert (d.toISOString() == "1968-12-01T01:00:00.000Z");
+
+d = new Date("1970-01-31T01:00:00.000Z");
+d.setFullYear(1971);
+assert (d.toISOString() == "1971-01-31T01:00:00.000Z");
+
 /* Without argument */
 d = new Date();
 assert (isNaN (d.setTime()));