我很感兴趣如何把它放在循环中,以便获得cpu用来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;
}
尽管我输入了哪个数字测试,它在终端上总是写为零,我很清楚为什么因为 10 甚至 20 和 100 操作对于现代计算机来说不是什么,但我怎么能点这样才能得到如果不是准确的答案,近似值至少?非常感谢