我有一个简单的基准,用于演示busywait线程的性能。它以两种模式运行:第一种简单地依次获取两个时间点,第二种通过向量进行迭代并测量迭代的持续时间。我看到Clock :: now()的两个顺序调用平均花费大约50纳秒,而向量的一次平均迭代花费大约100纳秒。但是有时执行这些操作会有很大的延迟:第一种情况大约需要50微秒,第二种情况大约需要10毫秒(!)。
测试在单个隔离的内核上运行,因此不会发生上下文切换。我还在程序的开头调用mlockall,所以我认为页面错误不会影响性能。
还应用了以下其他优化:
内核启动参数:intel_idle.max_cstate = 0 idle = halt irqaffinity = 0,14 isolcpus = 4-13,16-27 pti = off spectre_v2 = off audit = 0 selinux = 0 nmi_watchdog = 0 nosoftlockup = 0 rcu_nocb_poll rcu_nocbs = 19-20 nohz_full = 19-20; rcu [^ c]内核线程移至内部管理CPU内核0; 网卡RxTx队列移至管理CPU核心0; 写回内核工作队列移至内部CPU内核0; transparent_hugepage已禁用; 英特尔CPU超线程已禁用; 不使用交换文件/分区。 环境:
System details:
Default Archlinux kernel:
5.1.9-arch1-1-ARCH #1 SMP PREEMPT Tue Jun 11 16:18:09 UTC 2019 x86_64 GNU/Linux
that has following PREEMPT and HZ settings:
CONFIG_HZ_300=y
CONFIG_HZ=300
CONFIG_PREEMPT=y
Hardware details:
RAM: 256GB
CPU(s): 28
On-line CPU(s) list: 0-27
Thread(s) per core: 1
Core(s) per socket: 14
Socket(s): 2
NUMA node(s): 2
Vendor ID: GenuineIntel
CPU family: 6
Model: 79
Model name: Intel(R) Xeon(R) CPU E5-2690 v4 @ 2.60GHz
Stepping: 1
CPU MHz: 3200.011
CPU max MHz: 3500.0000
CPU min MHz: 1200.0000
BogoMIPS: 5202.68
Virtualization: VT-x
L1d cache: 32K
L1i cache: 32K
L2 cache: 256K
L3 cache: 35840K
NUMA node0 CPU(s): 0-13
NUMA node1 CPU(s): 14-27
示例代码:
struct TData
{
std::vector<char> Data;
TData() = default;
TData(size_t aSize)
{
for (size_t i = 0; i < aSize; ++i)
{
Data.push_back(i);
}
}
};
using TBuffer = std::vector<TData>;
TData DoMemoryOperation(bool aPerform, const TBuffer& aBuffer, size_t& outBufferIndex)
{
if (!aPerform)
{
return TData {};
}
const TData& result = aBuffer[outBufferIndex];
if (++outBufferIndex == aBuffer.size())
{
outBufferIndex = 0;
}
return result;
}
void WarmUp(size_t aCyclesCount, bool aPerform, const TBuffer& aBuffer)
{
size_t bufferIndex = 0;
for (size_t i = 0; i < aCyclesCount; ++i)
{
auto data = DoMemoryOperation(aPerform, aBuffer, bufferIndex);
}
}
void TestCycle(size_t aCyclesCount, bool aPerform, const TBuffer& aBuffer, Measurings& outStatistics)
{
size_t bufferIndex = 0;
for (size_t i = 0; i < aCyclesCount; ++i)
{
auto t1 = std::chrono::steady_clock::now();
{
auto data = DoMemoryOperation(aPerform, aBuffer, bufferIndex);
}
auto t2 = std::chrono::steady_clock::now();
auto diff = std::chrono::duration_cast<std::chrono::nanoseconds>(t2 - t1).count();
outStatistics.AddMeasuring(diff, t2);
}
}
int Run(int aCpu, size_t aDataSize, size_t aBufferSize, size_t aCyclesCount, bool aAllocate, bool aPerform)
{
if (mlockall(MCL_CURRENT | MCL_FUTURE))
{
throw std::runtime_error("mlockall failed");
}
std::cout << "Test parameters"
<< ":\ndata size=" << aDataSize
<< ",\nnumber of elements=" << aBufferSize
<< ",\nbuffer size=" << aBufferSize * aDataSize
<< ",\nnumber of cycles=" << aCyclesCount
<< ",\nallocate=" << aAllocate
<< ",\nperform=" << aPerform
<< ",\nthread ";
SetCpuAffinity(aCpu);
TBuffer buffer;
if (aPerform)
{
buffer.resize(aBufferSize);
std::fill(buffer.begin(), buffer.end(), TData { aDataSize });
}
WaitForKey();
std::cout << "Running..."<< std::endl;
WarmUp(aBufferSize * 2, aPerform, buffer);
Measurings statistics;
TestCycle(aCyclesCount, aPerform, buffer, statistics);
statistics.Print(aCyclesCount);
WaitForKey();
if (munlockall())
{
throw std::runtime_error("munlockall failed");
}
return 0;
}
并收到以下结果:首先:
StandaloneTests --run_test=MemoryAccessDelay --cpu=19 --data-size=280 --size=67108864 --count=1000000000 --allocate=1 --perform=0
Test parameters:
data size=280,
number of elements=67108864,
buffer size=18790481920,
number of cycles=1000000000,
allocate=1,
perform=0,
thread 14056 on cpu 19
Statistics: min: 16: max: 18985: avg: 18
0 - 10 : 0 (0 %): -
10 - 100 : 999993494 (99 %): min: 40: max: 117130: avg: 40
100 - 1000 : 946 (0 %): min: 380: max: 506236837: avg: 43056598
1000 - 10000 : 5549 (0 %): min: 56876: max: 70001739: avg: 7341862
10000 - 18985 : 11 (0 %): min: 1973150818: max: 14060001546: avg: 3644216650
第二:
StandaloneTests --run_test=MemoryAccessDelay --cpu=19 --data-size=280 --size=67108864 --count=1000000000 --allocate=1 --perform=1
Test parameters:
data size=280,
number of elements=67108864,
buffer size=18790481920,
number of cycles=1000000000,
allocate=1,
perform=1,
thread 3264 on cpu 19
Statistics: min: 36: max: 4967479: avg: 48
0 - 10 : 0 (0 %): -
10 - 100 : 964323921 (96 %): min: 60: max: 4968567: avg: 74
100 - 1000 : 35661548 (3 %): min: 122: max: 4972632: avg: 2023
1000 - 10000 : 14320 (0 %): min: 1721: max: 33335158: avg: 5039338
10000 - 100000 : 130 (0 %): min: 10010533: max: 1793333832: avg: 541179510
100000 - 1000000 : 0 (0 %): -
1000000 - 4967479 : 81 (0 %): min: 508197829: max: 2456672083: avg: 878824867
有什么想法会造成如此大的延误的原因是什么?如何进行调查?
在:
TData DoMemoryOperation(bool aPerform, const TBuffer& aBuffer, size_t& outBufferIndex);
它返回一个std::vector 按值。这涉及内存分配和数据复制。内存分配可以执行syscall(brk或mmap),并且与内存映射相关的syscall速度慢是众所周知的。
当计时包括系统调用时,不能指望低方差。
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