为什么Skylake比Broadwell-E在单线程内存吞吐量方面要好得多？

Question

我们有一个简单的内存吞吐量基准。对于大块内存，它所做的只是重复记忆。

在几台不同的机器上查看结果（针对64位编译），Skylake机器的性能明显优于Broadwell-E，保持OS（Win10-64），处理器速度和RAM速度（DDR4-2133）不变。我们不是说几个百分点，而是大约2个因子.Skylake配置为双通道，Broadwell-E的结果不会因双/三/四通道而变化。

任何想法为什么会这样？随后的代码在VS2015的Release中编译，并报告完成每个memcpy的平均时间：

64位：Skylake为2.2ms，Broadwell-E为4.5ms

32位：Skylake为2.2ms，Broadwell-E为3.5ms。

通过利用多个线程，我们可以在四通道Broadwell-E构建上获得更大的内存吞吐量，这很不错，但是看到单线程内存访问的这种巨大差异令人沮丧。为什么差异如此显着的任何想法？

我们还使用了各种基准测试软件，他们验证了这个简单示例所展示的内容 - 单线程内存吞吐量在Skylake上更好。

#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();
}

Answer 1

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