测量c ++中popcount函数的时间

Question

我感兴趣的是如何把它放在循环中,以便获得cpu执行每个不同操作的实时时间

#include<iostream>
#include<cstdlib>
#include<time.h>

using namespace std;
typedef unsigned __int64 uint64;
const uint64 m1=0x5555555555555555;
const uint64 m2=0x3333333333333333;
const uint64 m4=0x0f0f0f0f0f0f0f0f;
const uint64 m8=0x00ff00ff00ff00ff;
const uint64 m16=0x0000ffff0000ffff;
const uint64 m32=0x00000000ffffffff;
const uint64 hff=0xffffffffffffffff;
const uint64 h01=0x0101010101010101;

uint64 popcount_1(uint64 x)
{
    x=(x&m1)+((x>>1)&m1);
    x=(x&m2)+((x>>2)&m2);
    x=(x&m4)+((x>>4)&m4);
    x=(x&m8)+((x>>8)&m8);
    x=(x&m16)+((x>>16)&m16);
    x=(x&m32)+((x>>32)&m32);
    return (uint64)x;
}

//This uses fewer arithmetic operations than any other known
//implementation on machines with slow multiplication.
//It uses 17 arithmetic operations.
int popcount_2(uint64 x)
{
    x-=(x>>1)&m1;//put count of each 2 bits into those 2 bits
    x=(x&m2)+((x>>2)&m2);//put count of each 4 bits into those 4 bits
    x=(x+(x>>4))&m4; //put count of each 8 bits into those 8 bits
    x+=x>>8;//put count of each 16 bits into their lowest 8 bits
    x+=x>>16;
    x+=x>>32;
    return x&0x7f;
}
////This uses fewer arithmetic operations than any other known
//implementation on machines with fast multiplication.
//It uses 12 arithmetic operations, one of which is a multiply.
int popcount_3(uint64 x)
{
    x-=(x>>1)&m1;
    x=(x&m2)+((x>>2)&m2);
    x=(x+(x>>4))&m4;
    return (x*h01)>>56;
}
uint64 popcount_4(uint64 x)
{
    uint64  count;
    for(count=0; x; count++)
    {
        x&=x-1;
    }
    return count;
}
uint64 random()
{
    uint64 r30=RAND_MAX*rand()+rand();
    uint64 s30=RAND_MAX*rand()+rand();
    uint64  t4=rand()&0xf;
    uint64 res=(r30<<34 )+(s30<<4)+t4;
    return res;
}
int main()
{
    int testnum;
    while (true)
    {
        cout<<"enter number of test "<<endl;
        cin>>testnum;
        uint64 x= random();
        switch(testnum)
        {
            case 1: {
                    clock_t start=clock();
                    popcount_1(x);
                    clock_t end=clock();
                    cout<<"execution time of first method"<<start-end<<" "<<endl;
                }
                break;
            case 2: {
                    clock_t start=clock();
                    popcount_2(x);
                    clock_t end=clock();
                    cout<<"execution time of first method"<<start-end<<" "<<endl;
                }
                break;
            case 3: {
                    clock_t start=clock();
                    popcount_3(x);
                    clock_t end=clock();
                    cout<<"execution time of first method"<<start-end<<" "<<endl;
                }
                break;
            case 4: {
                    clock_t start=clock();
                    popcount_4(x);
                    clock_t end=clock();
                    cout<<"execution time of first method"<<start-end<<" "<<endl;
                }
                break;
            default:
                cout<<"it is not correct number "<<endl;
                break;
        }
    }
    return 0;
}

它写在终端总是零,尽管我进入了哪个数字测试,我很清楚为什么因为10或甚至20和100操作对现代计算机来说都不是什么,但我怎么能点到这样得到如果不是精确答案,近似于至少？非常感谢

Answer 1

只需重复所有测试.

以下每次测试重复1 Mio(1024*1024) 2 ^ 25次.您可能希望将时间除以获得以纳秒为单位的时间,但为了进行比较,总数将很好(并且更容易阅读).

int main()
{
    int testnum;
    while (true)
    {
        cout<<"enter number of test "<<endl;
        cin>>testnum;
        uint64 x= random();

        clock_t start=clock();
        switch(testnum)
        {
            case 1: for(unsigned long it=0; it<=(1ul<<26); ++it) popcount_1(x); break;
            case 2: for(unsigned long it=0; it<=(1ul<<26); ++it) popcount_2(x); break;
            case 3: for(unsigned long it=0; it<=(1ul<<26); ++it) popcount_3(x); break;
            case 4: for(unsigned long it=0; it<=(1ul<<26); ++it) popcount_4(x); break;
            default:
                cout<<"it is not correct number "<<endl;
                break;
        }
        clock_t end=clock();
        cout<<"execution time of method " << testnum << ": " << (end-start) <<" "<<endl;
    }
    return 0;
}

注意也固定 start-end为(end-start):)