1 files changed, 72 insertions, 0 deletions
diff --git a/src/algorithms/bits/MinMax.java b/src/algorithms/bits/MinMax.java
new file mode 100644
index 0000000..624ea57
--- /dev/null
+++ b/src/algorithms/bits/MinMax.java
@@ -0,0 +1,72 @@
+/**
+ * Copyright © 2014  Mattias Andrée (maandree@member.fsf.org)
+ * 
+ * This program is free software: you can redistribute it and/or modify
+ * it under the terms of the GNU Affero General Public License as published by
+ * the Free Software Foundation, either version 3 of the License, or
+ * (at your option) any later version.
+ * 
+ * This program is distributed in the hope that it will be useful,
+ * but WITHOUT ANY WARRANTY; without even the implied warranty of
+ * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+ * GNU Affero General Public License for more details.
+ * 
+ * You should have received a copy of the GNU Affero General Public License
+ * along with this program.  If not, see <http://www.gnu.org/licenses/>.
+ */
+package algorithms.bits;
+
+
+/**
+ * Compute the minimum or maximum of two integers
+ */
+public class MinMax
+{
+£<for T_S in char_15 byte_7 short_15 int_31 long_63; do
+  T=${T_S%_*}
+£>S=${T_S#*_}
+    /**
+     * Compute the minimum of two integers
+     * 
+     * @param   a  One of the integers
+     * @param   b  The other of the integers
+     * @return     The minimum of the two integers
+     */
+    public static £{T} min(£{T} a, £{T} b)
+    {
+	return (£{T})(b ^ (£{T})((a ^ b) & ((a - b) >> £{S})));
+	/* In C you can also do:
+	 *     b ^ ((a ^ b) & -(a < b))
+	 */
+	/*
+	 * Other version that can be faster:
+	 *     (a < b) ? a : b
+	 *     Assuming you now that INT_MIN <= x - y <= INT_MAX:
+	 *         b + ((a - b) & ((a - b) >> (sizeof(£{T}) * CHAR_BITS - 1)))
+	 */
+    }
+    
+    /**
+     * Compute the maximum of two integers
+     * 
+     * @param   a  One of the integers
+     * @param   b  The other of the integers
+     * @return     The maximum of the two integers
+     */
+    public static £{T} max(£{T} a, £{T} b)
+    {
+	return (£{T})(a ^ (£{T})((a ^ b) & ((a - b) >> £{S})));
+	/* In C you can also do:
+	 *    a ^ ((a ^ b) & -(a < b))
+	 */
+	/*
+	 * Other version that can be faster:
+	 *     (a < b) ? b : a
+	 *     Assuming you now that INT_MIN <= x - y <= INT_MAX:
+	 *         a - ((a - b) & ((a - b) >> (sizeof(£{T}) * CHAR_BITS - 1)))
+	 */
+    }
+    
+£>done
+}
+