数字比较花费的时间
比较次数：log(2^28) passes over 2^28 numbers = ~2^33 comparisons
1/2的比较会误判，因此 2^32 mispredicts * 5ns/mispredictt = 21 s
内存拷贝(顺序访问)
2^30 bytes * 28 passes = 28GB
Mermory bindwidth ~4GB/s
大约为 28GB/4Gb = 7s
总共需要大约30s的时间来排序1GB的数据。
下面我们来写测试一下实际的运行时间：
#include <iostream>
#include <vector>
#include <algorithm>
#include <sys/time.h>
#include <time.h>
#include <stdint.h>
using namespace std;
double get_time() {
struct timeval tv;
gettimeofday(&tv, NULL);
return tv.tv_sec + tv.tv_usec / 1000000.0;
}
int main() {
vector<int> array(1024 * 1024 * 1024 / sizeof(int32_t));
srand(unsigned(get_time()));
for (size_t i = 0; i < array.size(); i++) {
array[i] = rand();
}
double start = get_time();
sort(array.begin(), array.end());
double end = get_time();
cout << "time cost: " << end - start << endl;
return 0;
}
以O2的参数编译并且运行
g++ -O2 && ./a.out time cost: 34.3306
附录：
Numbers Everyone Should Know
| L1 cache reference | 0.5 ns |
| Branch mispredict | 5 ns |
| L2 cache reference | 7 ns |
| Mutex lock/unlock | 100 ns |
| Main memory reference | 100 ns |
| Compress 1K bytes with Zippy | 10,000 ns |
| Send 1K bytes over 1 Gbps network | 10,000 ns |
| Read 1 MB sequentially from memory | 250,000 ns |
| Round trip within same datacenter | 500,000 ns |
| Disk seek | 10,000,000 ns |
| Read 1 MB sequentially from network | 10,000,000 ns |
| Read 1 MB sequentially from disk | 30,000,000 ns |
| Send packet CA->Netherlands->CA | 150,000,000 ns |
