L1缓存小姐的费用是多少？
编辑 : 出于参考目的(如果有人碰到这个问题) ，Igor Ostrovsky 写了一个关于缓存丢失的 很棒的帖子。它讨论了几个不同的问题，并显示了示例数字。 结束编辑
我对 <long story goes here>进行了一些测试，想知道性能差异是否是由于内存缓存丢失造成的。下面的代码演示了这个问题，并将其归结为关键的计时部分。下面的代码有两个循环，它们以随机顺序访问内存，然后以升序地址顺序访问内存。
我在 XP 机器(用 VS2005: cl/O2编译)和 Linux 机器(gcc-Os)上运行它。两者产生了相似的时间。这些时间以毫秒为单位。我相信所有的循环都在运行，并且没有优化(否则它会“立即”运行)。
*** Testing 20000 nodes Total Ordered Time: 888.822899 Total Random Time: 2155.846268
这些数字说得通吗？这种差异主要是由于 L1缓存丢失还是其他原因造成的？存在20,000 ^ 2个内存访问，如果每个访问都是缓存错过，那么每次错过大约是3.2纳秒。我测试的 XP (P4)机器是3.2 GHz，我怀疑(但不知道)有一个32KB L1缓存和512KB L2。对于20,000个条目(80KB) ，我假设不会有大量的 L2错过。这就是 (3.2*10^9 cycles/second) * 3.2*10^-9 seconds/miss) = 10.1 cycles/miss。我觉得挺高的。也许不是，也许我数学不好。我试过用 VTune 测量缓存未命中率但我得到了一个 BSOD。现在我无法让它连接到许可证服务器(grrrr)。
typedef struct stItem
{
long lData;
//char acPad[20];
} LIST_NODE;
#if defined( WIN32 )
void StartTimer( LONGLONG *pt1 )
{
QueryPerformanceCounter( (LARGE_INTEGER*)pt1 );
}
void StopTimer( LONGLONG t1, double *pdMS )
{
LONGLONG t2, llFreq;
QueryPerformanceCounter( (LARGE_INTEGER*)&t2 );
QueryPerformanceFrequency( (LARGE_INTEGER*)&llFreq );
*pdMS = ((double)( t2 - t1 ) / (double)llFreq) * 1000.0;
}
#else
// doesn't need 64-bit integer in this case
void StartTimer( LONGLONG *pt1 )
{
// Just use clock(), this test doesn't need higher resolution
*pt1 = clock();
}
void StopTimer( LONGLONG t1, double *pdMS )
{
LONGLONG t2 = clock();
*pdMS = (double)( t2 - t1 ) / ( CLOCKS_PER_SEC / 1000 );
}
#endif
long longrand()
{
#if defined( WIN32 )
// Stupid cheesy way to make sure it is not just a 16-bit rand value
return ( rand() << 16 ) | rand();
#else
return rand();
#endif
}
// get random value in the given range
int randint( int m, int n )
{
int ret = longrand() % ( n - m + 1 );
return ret + m;
}
// I think I got this out of Programming Pearls (Bentley).
void ShuffleArray
(
long *plShuffle, // (O) return array of "randomly" ordered integers
long lNumItems // (I) length of array
)
{
long i;
long j;
long t;
for ( i = 0; i < lNumItems; i++ )
plShuffle[i] = i;
for ( i = 0; i < lNumItems; i++ )
{
j = randint( i, lNumItems - 1 );
t = plShuffle[i];
plShuffle[i] = plShuffle[j];
plShuffle[j] = t;
}
}
int main( int argc, char* argv[] )
{
long *plDataValues;
LIST_NODE *pstNodes;
long lNumItems = 20000;
long i, j;
LONGLONG t1; // for timing
double dms;
if ( argc > 1 && atoi(argv[1]) > 0 )
lNumItems = atoi( argv[1] );
printf( "\n\n*** Testing %u nodes\n", lNumItems );
srand( (unsigned int)time( 0 ));
// allocate the nodes as one single chunk of memory
pstNodes = (LIST_NODE*)malloc( lNumItems * sizeof( LIST_NODE ));
assert( pstNodes != NULL );
// Create an array that gives the access order for the nodes
plDataValues = (long*)malloc( lNumItems * sizeof( long ));
assert( plDataValues != NULL );
// Access the data in order
for ( i = 0; i < lNumItems; i++ )
plDataValues[i] = i;
StartTimer( &t1 );
// Loop through and access the memory a bunch of times
for ( j = 0; j < lNumItems; j++ )
{
for ( i = 0; i < lNumItems; i++ )
{
pstNodes[plDataValues[i]].lData = i * j;
}
}
StopTimer( t1, &dms );
printf( "Total Ordered Time: %f\n", dms );
// now access the array positions in a "random" order
ShuffleArray( plDataValues, lNumItems );
StartTimer( &t1 );
for ( j = 0; j < lNumItems; j++ )
{
for ( i = 0; i < lNumItems; i++ )
{
pstNodes[plDataValues[i]].lData = i * j;
}
}
StopTimer( t1, &dms );
printf( "Total Random Time: %f\n", dms );
}
最佳答案