问题
Edit: For reference purposes (if anyone stumbles across this question), Igor Ostrovsky wrote a great post about cache misses. It discusses several different issues and shows example numbers. End Edit
I did some testing <long story goes here> and am wondering if a performance difference is due to memory cache misses. The following code demonstrates the issue and boils it down to the critical timing portion. The following code has a couple of loops that visit memory in random order and then in ascending address order.
I ran it on an XP machine (compiled with VS2005: cl /O2) and on a Linux box (gcc –Os). Both produced similar times. These times are in milliseconds. I believe all loops are running and are not optimized out (otherwise it would run “instantly”).
*** Testing 20000 nodes Total Ordered Time: 888.822899 Total Random Time: 2155.846268
Do these numbers make sense? Is the difference primarily due to L1 cache misses or is something else going on as well? There are 20,000^2 memory accesses and if every one were a cache miss, that is about 3.2 nanoseconds per miss. The XP (P4) machine I tested on is 3.2GHz and I suspect (but don’t know) has a 32KB L1 cache and 512KB L2. With 20,000 entries (80KB), I assume there is not a significant number of L2 misses. So this would be (3.2*10^9 cycles/second) * 3.2*10^-9 seconds/miss) = 10.1 cycles/miss. That seems high to me. Maybe it’s not, or maybe my math is bad. I tried measuring cache misses with VTune, but I got a BSOD. And now I can’t get it to connect to the license server (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 );
}
回答1:
While I can't offer an answer to whether or not the numbers make sense (I'm not well versed in the cache latencies, but for the record ~10 cycle L1 cache misses sounds about right), I can offer you Cachegrind as a tool to help you actually see the differences in cache performance between your 2 tests.
Cachegrind is a Valgrind tool (the framework that powers the always-lovely memcheck) which profiles cache and branch hits/misses. It will give you an idea of how many cache hits/misses you are actually getting in your program.
回答2:
Here is an attempt to provide insight into the relative cost of cache misses by analogy with baking chocolate chip cookies ...
Your hands are your registers. It takes you 1 second to drop chocolate chips into the dough.
The kitchen counter is your L1 cache, twelve times slower than registers. It takes 12 x 1 = 12 seconds to step to the counter, pick up the bag of walnuts, and empty some into your hand.
The fridge is your L2 cache, four times slower than L1. It takes 4 x 12 = 48 seconds to walk to the fridge, open it, move last night's leftovers out of the way, take out a carton of eggs, open the carton, put 3 eggs on the counter, and put the carton back in the fridge.
The cupboard is your L3 cache, three times slower than L2. It takes 3 x 48 = 2 minutes and 24 seconds to take three steps to the cupboard, bend down, open the door, root around to find the baking supply tin, extract it from the cupboard, open it, dig to find the baking powder, put it on the counter and sweep up the mess you spilled on the floor.
And main memory? That's the corner store, 5 times slower than L3. It takes 5 x 2:24 = 12 minutes to find your wallet, put on your shoes and jacket, dash down the street, grab a litre of milk, dash home, take off your shoes and jacket, and get back to the kitchen.
Note that all these accesses are constant complexity -- O(1) -- but the differences between them can have a huge impact on performance. Optimizing purely for big-O complexity is like deciding whether to add chocolate chips to the batter 1 at a time or 10 at a time, but forgetting to put them on your grocery list.
Moral of the story: Organize your memory accesses so the CPU has to go for groceries as rarely as possible.
Numbers were taken from the CPU Cache Flushing Fallacy blog post, which indicates that for a particular 2012-era Intel processor, the following is true:
- register access = 4 instructions per cycle
- L1 latency = 3 cycles (12 x register)
- L2 latency = 12 cycles (4 x L1, 48 x register)
- L3 latency = 38 cycles (3 x L2, 12 x L1, 144 x register)
- DRAM latency = 65 ns = 195 cycles on a 3 GHz CPU (5 x L3, 15 x L2, 60 x L1, 720 x register)
The Gallery of Processor Cache Effects also makes good reading on this topic.
回答3:
3.2ns for an L1 cache miss is entirely plausible. For comparison, on one particular modern multicore PowerPC CPU, an L1 miss is about 40 cycles -- a little longer for some cores than others, depending on how far they are from the L2 cache (yes really). An L2 miss is at least 600 cycles.
Cache is everything in performance; CPUs are so much faster than memory now that you're really almost optimizing for the memory bus instead of the core.
回答4:
Well yeah that does look like it will mainly be L1 cache misses.
10 cycles for an L1 cache miss does sound about reasonable, probably a little on the low side.
A read from RAM is going to take of the order of 100s or may be even 1000s (Am too tired to attempt to do the maths right now ;)) of cycles so its still a huge win over that.
回答5:
If you plan on using cachegrind, please note that it is a cache hit/miss simulator only. It won't always be accurate. For example: if you access some memory location, say 0x1234 in a loop 1000 times, cachegrind will always show you that there was only one cache miss (the first access) even if you have something like:
clflush 0x1234 in your loop.
On x86, this will cause all 1000 cache misses.
回答6:
Some numbers for a 3.4GHz P4 from a Lavalys Everest run:
- the L1 dcache is 8K (cacheline 64 bytes)
- L2 is 512K
- L1 fetch latency is 2 cycles
- L2 fetch latency is about double what you are seeing: 20 cycles
More here: http://www.freeweb.hu/instlatx64/GenuineIntel0000F25_P4_Gallatin_MemLatX86.txt
(for the latencies look at the bottom of the page)
回答7:
It's difficult to say anything for sure without a lot more testing, but in my experience that scale of difference definitely can be attributed to the CPU L1 and/or L2 cache, especially in a scenario with randomized access. You could probably make it even worse by ensuring that each access is at least some minimum distance from the last.
回答8:
The easiest thing to do is to take a scaled photograph of the target cpu and physically measure the distance between the core and the level-1 cache. Multiply that distance by the distance electrons can travel per second in copper. Then figure out how many clock-cycles you can have in that same time. That's the minimum number of cpu cycles you'll waste on a L1 cache miss.
You can also work out the minimum cost of fetching data from RAM in terms of the number of CPU cycles wasted in the same way. You might be amazed.
Notice that what you're seeing here definitely has something to do with cache-misses (be it L1 or both L1 and L2) because normally the cache will pull out data on the same cache line once you access anything on that cache-line requiring less trips to RAM.
However, what you're probably also seeing is the fact that RAM (even though it's calls Random Access Memory) still preferres linear memory access.
来源：https://stackoverflow.com/questions/1126529/what-is-the-cost-of-an-l1-cache-miss