Instead of shifting the XOR value right and comparing it to 1, which
roughly requires 2 sequential instructions, better test if the XOR has
any bit above the current bit, which means any bit set among those
strictly higher, or in other words that XOR & (-bit << 1) is non-zero.
This is one less instruction in the fast path and gives another nice
performance gain on random keys (in million lookups/s):
eb32 1k: 33.17 -> 37.30 +12.5%
10k: 15.74 -> 17.08 +8.51%
100k: 8.00 -> 9.00 +12.5%
eb64 1k: 34.40 -> 38.10 +10.8%
10k: 16.17 -> 17.10 +5.75%
100k: 8.38 -> 8.87 +5.85%
This is ebtree commit
c942a2771758eed4f4584fe23cf2914573817a6b.
return node;
}
- if ((y >> node_bit) >= EB_NODE_BRANCHES)
+ if (y & -(z << 1))
return NULL; /* no more common bits */
}
}
return node;
}
- if ((y >> node_bit) >= EB_NODE_BRANCHES)
+ if (y & -(z << 1))
return NULL; /* no more common bits */
}
}
return node;
}
- if ((y >> node->node.bit) >= EB_NODE_BRANCHES)
+ if (y & -(z << 1))
return NULL; /* no more common bits */
}
}
return node;
}
- if ((y >> node->node.bit) >= EB_NODE_BRANCHES)
+ if (y & -(z << 1))
return NULL; /* no more common bits */
}
}