+/*
+ * given 6 bytes, view them as 8 6-bit numbers and return the XOR of those
+ * the code below is about seven times as fast as the original code
+ *
+ * The original code was:
+ *
+ * u32 xor( u32 a, u32 b ) { return ( ( a && ! b ) || ( ! a && b ) ); }
+ *
+ * #define XOR8( a, b, c, d, e, f, g, h ) \
+ * xor( a, xor( b, xor( c, xor( d, xor( e, xor( f, xor( g, h ) ) ) ) ) ) )
+ * #define DA( a, bit ) ( ( (u8) a[bit/8] ) & ( (u8) ( 1 << bit%8 ) ) )
+ *
+ * hash = XOR8( DA(a,0), DA(a, 6), DA(a,12), DA(a,18), DA(a,24), DA(a,30), DA(a,36), DA(a,42) );
+ * hash |= XOR8( DA(a,1), DA(a, 7), DA(a,13), DA(a,19), DA(a,25), DA(a,31), DA(a,37), DA(a,43) ) << 1;
+ * hash |= XOR8( DA(a,2), DA(a, 8), DA(a,14), DA(a,20), DA(a,26), DA(a,32), DA(a,38), DA(a,44) ) << 2;
+ * hash |= XOR8( DA(a,3), DA(a, 9), DA(a,15), DA(a,21), DA(a,27), DA(a,33), DA(a,39), DA(a,45) ) << 3;
+ * hash |= XOR8( DA(a,4), DA(a,10), DA(a,16), DA(a,22), DA(a,28), DA(a,34), DA(a,40), DA(a,46) ) << 4;
+ * hash |= XOR8( DA(a,5), DA(a,11), DA(a,17), DA(a,23), DA(a,29), DA(a,35), DA(a,41), DA(a,47) ) << 5;
+ *
+ */
+static inline u32 TLan_HashFunc( const u8 *a )