[英]How can std::lock_guard be faster than std::mutex::lock()?
I was arguing with a colleague, about lock_guard, and he proposed that lock_guard is problably slower than mutex::lock() / mutex::unlock() due to the cost of instantiate and unistantiate the class lock_guard.我和一位同事争论 lock_guard,他提出 lock_guard 可能比 mutex::lock() / mutex::unlock() 慢,原因是实例化和非实例化类 lock_guard 的成本。
Then I created this simple test and, surprisely, the version with lock_guard is almost two times faster than the version with mutex::lock() / mutex::unlock()然后我创建了这个简单的测试,令人惊讶的是,带有 lock_guard 的版本几乎比带有 mutex::lock() / mutex::unlock() 的版本快两倍
#include <iostream>
#include <mutex>
#include <chrono>
std::mutex m;
int g = 0;
void func1()
{
m.lock();
g++;
m.unlock();
}
void func2()
{
std::lock_guard<std::mutex> lock(m);
g++;
}
int main()
{
auto t = std::chrono::system_clock::now();
for (int i = 0; i < 1000000; i++)
{
func1();
}
std::cout << "Take: " << std::chrono::duration_cast<std::chrono::milliseconds>(std::chrono::system_clock::now() - t).count() << " ms" << std::endl;
t = std::chrono::system_clock::now();
for (int i = 0; i < 1000000; i++)
{
func2();
}
std::cout << "Take: " << std::chrono::duration_cast<std::chrono::milliseconds>(std::chrono::system_clock::now() - t).count() << " ms" << std::endl;
return 0;
}
The results on my machine:我机器上的结果:
Take: 41 ms
Take: 22 ms
Can someone clarify why and how this can be?有人可以澄清为什么以及如何做到这一点吗?
The release build produces the same result for both versions.发布版本对两个版本产生相同的结果。
The DEBUG
build shows ~33% longer time for func2
; DEBUG
构建显示func2
时间延长了约 33%; the difference I see in the disassembly that func2
uses __security_cookie
and invokes @_RTC_CheckStackVars@8
.我在反汇编中看到的区别是
func2
使用__security_cookie
并调用@_RTC_CheckStackVars@8
。
Are you timing DEBUG?你在定时调试吗?
EDIT: Additionally, while looking at RELEASE
disassembly, I noticed that mutex
methods were saved in two registries:编辑:此外,在查看
RELEASE
反汇编时,我注意到mutex
方法保存在两个注册表中:
010F104E mov edi,dword ptr [__imp___Mtx_lock (010F3060h)]
010F1054 xor esi,esi
010F1056 mov ebx,dword ptr [__imp___Mtx_unlock (010F3054h)]
and called the same way from both func1
and func2
:并从
func1
和func2
以相同的方式func2
:
010F1067 call edi
....
010F107F call ebx
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