We've got a simple memory throughput benchmark. All it does is memcpy repeatedly for a large block of memory.
Looking at the results (compiled for 64-bit) on a few different machines, Skylake machines do significantly better than Broadwell-E, keeping OS (Win10-64), processor speed, and RAM speed (DDR4-2133) the same. We're not talking a few percentage points, but rather a factor of about 2. Skylake is configured dual-channel, and the results for Broadwell-E don't vary for dual/triple/quad-channel.
Any ideas why this might be happening? The code that follows is compiled in Release in VS2015, and reports average time to complete each memcpy at:
64-bit: 2.2ms for Skylake vs 4.5ms for Broadwell-E
32-bit: 2.2ms for Skylake vs 3.5ms for Broadwell-E.
We can get greater memory throughput on a quad-channel Broadwell-E build by utilizing multiple threads, and that's nice, but to see such a drastic difference for single-threaded memory access is frustrating. Any thoughts on why the difference is so pronounced?
We've also used various benchmarking software, and they validate what this simple example shows - single-threaded memory throughput is way better on Skylake.
#include <memory>
#include <Windows.h>
#include <iostream>
//Prevent the memcpy from being optimized out of the for loop
_declspec(noinline) void MemoryCopy(void *destinationMemoryBlock, void *sourceMemoryBlock, size_t size)
{
memcpy(destinationMemoryBlock, sourceMemoryBlock, size);
}
int main()
{
const int SIZE_OF_BLOCKS = 25000000;
const int NUMBER_ITERATIONS = 100;
void* sourceMemoryBlock = malloc(SIZE_OF_BLOCKS);
void* destinationMemoryBlock = malloc(SIZE_OF_BLOCKS);
LARGE_INTEGER Frequency;
QueryPerformanceFrequency(&Frequency);
while (true)
{
LONGLONG total = 0;
LONGLONG max = 0;
LARGE_INTEGER StartingTime, EndingTime, ElapsedMicroseconds;
for (int i = 0; i < NUMBER_ITERATIONS; ++i)
{
QueryPerformanceCounter(&StartingTime);
MemoryCopy(destinationMemoryBlock, sourceMemoryBlock, SIZE_OF_BLOCKS);
QueryPerformanceCounter(&EndingTime);
ElapsedMicroseconds.QuadPart = EndingTime.QuadPart - StartingTime.QuadPart;
ElapsedMicroseconds.QuadPart *= 1000000;
ElapsedMicroseconds.QuadPart /= Frequency.QuadPart;
total += ElapsedMicroseconds.QuadPart;
max = max(ElapsedMicroseconds.QuadPart, max);
}
std::cout << "Average is " << total*1.0 / NUMBER_ITERATIONS / 1000.0 << "ms" << std::endl;
std::cout << "Max is " << max / 1000.0 << "ms" << std::endl;
}
getchar();
}
Question&Answers:
os 与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…