C++ 向量<；双倍>；比双倍还快*：为什么？
这里有一个循环，我用C++ 向量<；双倍>；比双倍还快*：为什么？,c++,arrays,optimization,vector,ode,C++,Arrays,Optimization,Vector,Ode,这里有一个循环，我用std:：vector和普通的double*尝试过 对于1000万个元素，向量版本在double*版本所需时间的80%左右持续运行；对于几乎任何N的值，vector都明显更快 纵观GCC STL源代码，我看不出std:：vector所做的事情本质上比double*惯用法所做的事情更有趣（即，使用普通的旧new[]，操作符[]取消对偏移量的引用）。我也这么说 你知道为什么矢量版本更快吗 Compiler: GCC 4.6.1 Example compile line: g++
std:：vectordouble*double*Nstd:：vectordouble*new[]操作符[]Compiler: GCC 4.6.1
Example compile line: g++ -Ofast -march=native -DNDEBUG \
-ftree-vectorizer-verbose=2 -o vector.bin \
vector.cpp -lrt
OS: CentOS 5
CPU: Opteron 8431
RAM: 128 GB
std:：vector#include <vector>
#include <iostream>
#include <boost/lexical_cast.hpp>
#include "util.h"
#include "rkck_params.h"
using namespace std;
int main( int argc, char* argv[] )
{
const size_t N = boost::lexical_cast<size_t>( argv[ 1 ] );
vector<double> y_old( N );
vector<double> y_new( N );
vector<double> y_err( N );
vector<double> k0( N );
vector<double> k1( N );
vector<double> k2( N );
vector<double> k3( N );
vector<double> k4( N );
vector<double> k5( N );
const double h = 0.5;
const timespec start = clock_tick();
for ( size_t i = 0 ; i < N ; ++i )
{
y_new[ i ] = y_old[ i ]
+ h
*(
rkck::c[ 0 ]*k0[ i ]
+ rkck::c[ 2 ]*k2[ i ]
+ rkck::c[ 3 ]*k3[ i ]
+ rkck::c[ 5 ]*k5[ i ]
);
y_err[ i ] = h
*(
rkck::cdiff[ 0 ]*k0[ i ]
+ rkck::cdiff[ 2 ]*k2[ i ]
+ rkck::cdiff[ 3 ]*k3[ i ]
+ rkck::cdiff[ 4 ]*k4[ i ]
+ rkck::cdiff[ 5 ]*k5[ i ]
);
}
const timespec stop = clock_tick();
const double total_time = seconds( start, stop );
// Output
cout << "vector\t" << N << "\t" << total_time << endl;
return 0;
}
rkck_参数h#ifndef RKCK_PARAMS_H
#define RKCK_PARAMS_H
namespace rkck
{
// C.f. $c_i$ in Ch. 16.2 of NR in C++, 2nd ed.
const double c[ 6 ]
= { 37.0/378.0,
0.0,
250.0/621.0,
125.0/594,
0.0,
512.0/1771.0 };
// C.f. $( c_i - c_i^* )$ in Ch. 16.2 of NR in C++, 2nd ed.
const double cdiff[ 6 ]
= { c[ 0 ] - 2825.0/27648.0,
c[ 1 ] - 0.0,
c[ 2 ] - 18575.0/48384.0,
c[ 3 ] - 13525.0/55296.0,
c[ 4 ] - 277.0/14336.0,
c[ 5 ] - 1.0/4.0 };
}
#endif
#ifndef UTIL_H
#define UTIL_H
#include <time.h>
#include <utility>
inline timespec clock_tick()
{
timespec tick;
clock_gettime( CLOCK_REALTIME, &tick );
return tick;
}
// \cite{www.guyrutenberg.com/2007/09/22/profiling-code-using-clock_gettime}
inline double seconds( const timespec& earlier, const timespec& later )
{
double seconds_diff = -1.0;
double nano_diff = -1.0;
if ( later.tv_nsec < earlier.tv_nsec )
{
seconds_diff = later.tv_sec - earlier.tv_sec - 1;
nano_diff = ( 1.0e9 + later.tv_nsec - earlier.tv_nsec )*1.0e-9;
}
else
{
seconds_diff = later.tv_sec - earlier.tv_sec;
nano_diff = ( later.tv_nsec - earlier.tv_nsec )*1.0e-9;
}
return seconds_diff + nano_diff;
}
#endif
util.h#ifndef RKCK_PARAMS_H
#define RKCK_PARAMS_H
namespace rkck
{
// C.f. $c_i$ in Ch. 16.2 of NR in C++, 2nd ed.
const double c[ 6 ]
= { 37.0/378.0,
0.0,
250.0/621.0,
125.0/594,
0.0,
512.0/1771.0 };
// C.f. $( c_i - c_i^* )$ in Ch. 16.2 of NR in C++, 2nd ed.
const double cdiff[ 6 ]
= { c[ 0 ] - 2825.0/27648.0,
c[ 1 ] - 0.0,
c[ 2 ] - 18575.0/48384.0,
c[ 3 ] - 13525.0/55296.0,
c[ 4 ] - 277.0/14336.0,
c[ 5 ] - 1.0/4.0 };
}
#endif
#ifndef UTIL_H
#define UTIL_H
#include <time.h>
#include <utility>
inline timespec clock_tick()
{
timespec tick;
clock_gettime( CLOCK_REALTIME, &tick );
return tick;
}
// \cite{www.guyrutenberg.com/2007/09/22/profiling-code-using-clock_gettime}
inline double seconds( const timespec& earlier, const timespec& later )
{
double seconds_diff = -1.0;
double nano_diff = -1.0;
if ( later.tv_nsec < earlier.tv_nsec )
{
seconds_diff = later.tv_sec - earlier.tv_sec - 1;
nano_diff = ( 1.0e9 + later.tv_nsec - earlier.tv_nsec )*1.0e-9;
}
else
{
seconds_diff = later.tv_sec - earlier.tv_sec;
nano_diff = ( later.tv_nsec - earlier.tv_nsec )*1.0e-9;
}
return seconds_diff + nano_diff;
}
#endif
\ifndef UTIL\u H
#定义UTIL_H
#包括
#包括
内联timespec时钟_tick（）
{
timespec滴答声；
时钟获取时间（时钟实时和滴答声）；
返回勾号；
}
//\cite{www.guyrutenberg.com/2007/09/22/profiling-code-using-clock_-gettime}
内联双秒（常数时间间隔和更早、常数时间间隔和更晚）
{
双秒_diff=-1.0；
双纳米差异=-1.0；
如果（later.tv\u nsecnew你跑过多次了吗，以不同的顺序？@Martinho:我非常同意，如果一个人的结果有任何意义的希望，真实的评测测试用例是至关重要的。可能重复@Martin:Is'这不是相反的问题吗？@MSalters:我假设人们足够聪明，能够颠倒问题以得到相同的答案。听起来不错：我在我的系统上复制OPs结果（vector快了50%），并将所有
new[]new[]（）N