提交 5eb7ac0c 编写于 作者: A antirez

Speedup hllRawSum() processing 8 bytes per iteration.

The internal HLL raw encoding used by PFCOUNT when merging multiple keys
is aligned to 8 bits (1 byte per register) so we can exploit this to
improve performances by processing multiple bytes per iteration.

In benchmarks the new code was several times faster with HLLs with many
registers set to zero, while no slowdown was observed with populated
HLLs.
上级 192a2132
......@@ -928,16 +928,24 @@ double hllSparseSum(uint8_t *sparse, int sparselen, double *PE, int *ezp, int *i
double hllRawSum(uint8_t *registers, double *PE, int *ezp) {
double E = 0;
int j, ez = 0;
unsigned long reg;
uint64_t *word = (uint64_t*) registers;
uint8_t *bytes;
for (j = 0; j < HLL_REGISTERS; j++) {
reg = registers[j];
if (reg == 0) {
ez++;
/* Increment E at the end of the loop. */
for (j = 0; j < HLL_REGISTERS/8; j++) {
if (*word == 0) {
ez += 8;
} else {
E += PE[reg]; /* Precomputed 2^(-reg[j]). */
bytes = (uint8_t*) word;
if (bytes[0]) E += PE[bytes[0]]; else ez++;
if (bytes[1]) E += PE[bytes[1]]; else ez++;
if (bytes[2]) E += PE[bytes[2]]; else ez++;
if (bytes[3]) E += PE[bytes[3]]; else ez++;
if (bytes[4]) E += PE[bytes[4]]; else ez++;
if (bytes[5]) E += PE[bytes[5]]; else ez++;
if (bytes[6]) E += PE[bytes[6]]; else ez++;
if (bytes[7]) E += PE[bytes[7]]; else ez++;
}
word++;
}
E += ez; /* 2^(-reg[j]) is 1 when m is 0, add it 'ez' times for every
zero register in the HLL. */
......
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