{
const auto __m = static_cast<unsigned>(month());
- // Excluding February, the last day of month __m is either 30 or 31 or,
- // in another words, it is 30 + b = 30 | b, where b is in {0, 1}.
+ // The result is unspecified if __m < 1 or __m > 12. Hence, assume
+ // 1 <= __m <= 12. For __m != 2, day() == 30 or day() == 31 or, in
+ // other words, day () == 30 | b, where b is in {0, 1}.
- // If __m in {1, 3, 4, 5, 6, 7}, then b is 1 if, and only if __m is odd.
- // Hence, b = __m & 1 = (__m ^ 0) & 1.
+ // If __m in {1, 3, 4, 5, 6, 7}, then b is 1 if, and only if, __m is
+ // odd. Hence, b = __m & 1 = (__m ^ 0) & 1.
- // If __m in {8, 9, 10, 11, 12}, then b is 1 if, and only if __m is even.
- // Hence, b = (__m ^ 1) & 1.
+ // If __m in {8, 9, 10, 11, 12}, then b is 1 if, and only if, __m is
+ // even. Hence, b = (__m ^ 1) & 1.
// Therefore, b = (__m ^ c) & 1, where c = 0, if __m < 8, or c = 1 if
// __m >= 8, that is, c = __m >> 3.
- // The above mathematically justifies this implementation whose
- // performance does not depend on look-up tables being on the L1 cache.
- return chrono::day{__m != 2 ? ((__m ^ (__m >> 3)) & 1) | 30
- : _M_y.is_leap() ? 29 : 28};
+ // Since 30 = (11110)_2 and __m <= 31 = (11111)_2, the "& 1" in b's
+ // calculation is unnecessary.
+
+ // The performance of this implementation does not depend on look-up
+ // tables being on the L1 cache.
+ return chrono::day{__m != 2 ? (__m ^ (__m >> 3)) | 30
+ : _M_y.is_leap() ? 29 : 28};
}
constexpr