我在看这个问题，因为我有同样的问题。我有一些想法要进行更严格的测试。
- 正如rodrigob指出的那样，循环顺序存在缺陷，因此您最初附加的代码中的任何结果都会产生误导性数据
- 此外，还有一些相当小的数组正在使用常量进行设置。编译器可能会优化循环，而实际上编译器不会知道数组的大小。数组的大小和迭代次数应该是运行时输入以防万一。
在 Mac 上，以下代码被配置为给出更有意义的答案。这里有4个测试。
#define BOOST_DISABLE_ASSERTS
#include "boost/multi_array.hpp"
#include <sys/time.h>
#include <stdint.h>
#include<string>
uint64_t GetTimeMs64()
{
struct timeval tv;
gettimeofday( &tv, NULL );
uint64_t ret = tv.tv_usec;
/* Convert from micro seconds (10^-6) to milliseconds (10^-3) */
ret /= 1000;
/* Adds the seconds (10^0) after converting them to milliseconds (10^-3) */
ret += ( tv.tv_sec * 1000 );
return ret;
}
void function1( const int X_SIZE, const int Y_SIZE, const int ITERATIONS )
{
double nativeMatrix1add[X_SIZE*Y_SIZE];
for( int x = 0 ; x < X_SIZE ; ++x )
{
for( int y = 0 ; y < Y_SIZE ; ++y )
{
nativeMatrix1add[y + ( x * Y_SIZE )] = rand();
}
}
// Create the native array
double* __restrict const nativeMatrix1p = new double[X_SIZE * Y_SIZE];
uint64_t startTime = GetTimeMs64();
for( int i = 0 ; i < ITERATIONS ; ++i )
{
for( int xy = 0 ; xy < X_SIZE*Y_SIZE ; ++xy )
{
nativeMatrix1p[xy] += nativeMatrix1add[xy];
}
}
uint64_t endTime = GetTimeMs64();
printf( "[Native Pointer] Elapsed time: %6.3f seconds\n", ( endTime - startTime ) / 1000.0 );
}
void function2( const int X_SIZE, const int Y_SIZE, const int ITERATIONS )
{
double nativeMatrix1add[X_SIZE*Y_SIZE];
for( int x = 0 ; x < X_SIZE ; ++x )
{
for( int y = 0 ; y < Y_SIZE ; ++y )
{
nativeMatrix1add[y + ( x * Y_SIZE )] = rand();
}
}
// Create the native array
double* __restrict const nativeMatrix1 = new double[X_SIZE * Y_SIZE];
uint64_t startTime = GetTimeMs64();
for( int i = 0 ; i < ITERATIONS ; ++i )
{
for( int x = 0 ; x < X_SIZE ; ++x )
{
for( int y = 0 ; y < Y_SIZE ; ++y )
{
nativeMatrix1[y + ( x * Y_SIZE )] += nativeMatrix1add[y + ( x * Y_SIZE )];
}
}
}
uint64_t endTime = GetTimeMs64();
printf( "[Native 1D Array] Elapsed time: %6.3f seconds\n", ( endTime - startTime ) / 1000.0 );
}
void function3( const int X_SIZE, const int Y_SIZE, const int ITERATIONS )
{
double nativeMatrix2add[X_SIZE][Y_SIZE];
for( int x = 0 ; x < X_SIZE ; ++x )
{
for( int y = 0 ; y < Y_SIZE ; ++y )
{
nativeMatrix2add[x][y] = rand();
}
}
// Create the native array
double nativeMatrix2[X_SIZE][Y_SIZE];
uint64_t startTime = GetTimeMs64();
for( int i = 0 ; i < ITERATIONS ; ++i )
{
for( int x = 0 ; x < X_SIZE ; ++x )
{
for( int y = 0 ; y < Y_SIZE ; ++y )
{
nativeMatrix2[x][y] += nativeMatrix2add[x][y];
}
}
}
uint64_t endTime = GetTimeMs64();
printf( "[Native 2D Array] Elapsed time: %6.3f seconds\n", ( endTime - startTime ) / 1000.0 );
}
void function4( const int X_SIZE, const int Y_SIZE, const int ITERATIONS )
{
boost::multi_array<double, 2> boostMatrix2add( boost::extents[X_SIZE][Y_SIZE] );
for( int x = 0 ; x < X_SIZE ; ++x )
{
for( int y = 0 ; y < Y_SIZE ; ++y )
{
boostMatrix2add[x][y] = rand();
}
}
// Create the native array
boost::multi_array<double, 2> boostMatrix( boost::extents[X_SIZE][Y_SIZE] );
uint64_t startTime = GetTimeMs64();
for( int i = 0 ; i < ITERATIONS ; ++i )
{
for( int x = 0 ; x < X_SIZE ; ++x )
{
for( int y = 0 ; y < Y_SIZE ; ++y )
{
boostMatrix[x][y] += boostMatrix2add[x][y];
}
}
}
uint64_t endTime = GetTimeMs64();
printf( "[Boost Array] Elapsed time: %6.3f seconds\n", ( endTime - startTime ) / 1000.0 );
}
int main( int argc, char* argv[] )
{
srand( time( NULL ) );
const int X_SIZE = std::stoi( argv[1] );
const int Y_SIZE = std::stoi( argv[2] );
const int ITERATIONS = std::stoi( argv[3] );
function1( X_SIZE, Y_SIZE, ITERATIONS );
function2( X_SIZE, Y_SIZE, ITERATIONS );
function3( X_SIZE, Y_SIZE, ITERATIONS );
function4( X_SIZE, Y_SIZE, ITERATIONS );
return 0;
}
一个只有一维数组，使用带有整数数学和双循环的 []
一个使用指针递增的具有相同一维数组的
多维 C 数组
提升多数组
所以从命令行运行，运行
./test_array xsize ysize iterations"
并且您可以很好地了解这些方法将如何执行。这是我使用以下编译器标志得到的结果：
g++4.9.2 -O3 -march=native -funroll-loops -mno-avx --fast-math -DNDEBUG -c -std=c++11
./test_array 51200 1 20000
[Native 1-Loop ] Elapsed time: 0.537 seconds
[Native 1D Array] Elapsed time: 2.045 seconds
[Native 2D Array] Elapsed time: 2.749 seconds
[Boost Array] Elapsed time: 1.167 seconds
./test_array 25600 2 20000
[Native 1-Loop ] Elapsed time: 0.531 seconds
[Native 1D Array] Elapsed time: 1.241 seconds
[Native 2D Array] Elapsed time: 1.631 seconds
[Boost Array] Elapsed time: 0.954 seconds
./test_array 12800 4 20000
[Native 1-Loop ] Elapsed time: 0.536 seconds
[Native 1D Array] Elapsed time: 1.214 seconds
[Native 2D Array] Elapsed time: 1.223 seconds
[Boost Array] Elapsed time: 0.798 seconds
./test_array 6400 8 20000
[Native 1-Loop ] Elapsed time: 0.540 seconds
[Native 1D Array] Elapsed time: 0.845 seconds
[Native 2D Array] Elapsed time: 0.878 seconds
[Boost Array] Elapsed time: 0.803 seconds
./test_array 3200 16 20000
[Native 1-Loop ] Elapsed time: 0.537 seconds
[Native 1D Array] Elapsed time: 0.661 seconds
[Native 2D Array] Elapsed time: 0.673 seconds
[Boost Array] Elapsed time: 0.708 seconds
./test_array 1600 32 20000
[Native 1-Loop ] Elapsed time: 0.532 seconds
[Native 1D Array] Elapsed time: 0.592 seconds
[Native 2D Array] Elapsed time: 0.596 seconds
[Boost Array] Elapsed time: 0.764 seconds
./test_array 800 64 20000
[Native 1-Loop ] Elapsed time: 0.546 seconds
[Native 1D Array] Elapsed time: 0.594 seconds
[Native 2D Array] Elapsed time: 0.606 seconds
[Boost Array] Elapsed time: 0.764 seconds
./test_array 400 128 20000
[Native 1-Loop ] Elapsed time: 0.536 seconds
[Native 1D Array] Elapsed time: 0.560 seconds
[Native 2D Array] Elapsed time: 0.564 seconds
[Boost Array] Elapsed time: 0.746 seconds
所以，我认为可以肯定地说 boost multi_array 的表现相当不错。没有什么比单循环评估更好的了，但是根据数组的维度，boost::multi_array 可能会击败具有双循环的标准 c 数组。