fprintf(stderr, "division by zero\n");
return false;
}
- *retval = lval / rval;
+
+ /*
+ * INT64_MIN / -1 is problematic, since the result
+ * can't be represented on a two's-complement machine.
+ * Some machines produce INT64_MIN, some produce zero,
+ * some throw an exception. We can dodge the problem
+ * by recognizing that division by -1 is the same as
+ * negation.
+ */
+ if (rval == -1)
+ {
+ *retval = -lval;
+
+ /* overflow check (needed for INT64_MIN) */
+ if (lval == PG_INT64_MIN)
+ {
+ fprintf(stderr, "bigint out of range\n");
+ return false;
+ }
+ }
+ else
+ *retval = lval / rval;
+
return true;
case '%':
fprintf(stderr, "division by zero\n");
return false;
}
- *retval = lval % rval;
+
+ /*
+ * Some machines throw a floating-point exception for
+ * INT64_MIN % -1. Dodge that problem by noting that
+ * any value modulo -1 is 0.
+ */
+ if (rval == -1)
+ *retval = 0;
+ else
+ *retval = lval % rval;
+
return true;
}