下面是重现错误的最小代码。
#include <iostream>
#include <mutex>
#include <vector>
class A {
std::mutex mutex;
public:
A(){};
};
int main()
{
std::vector<std::pair<std::string,A>> aa;
A a;
//aa.push_back(std::make_pair(std::string("aa"),A()));
//aa.push_back(std::make_pair(std::string("aa"),a));
aa.push_back(std::make_pair(std::string("aa"),std::move(a)));
}
以下是错误。
Microsoft(R)C / C ++优化编译器版本19.16.27026.1 for x64版权所有(C)Microsoft Corporation。版权所有。
> C:\Program Files (x86)\Microsoft Visual
> Studio\2017\Community\VC\Tools\MSVC\14.16.27023\include\xlocale(319):
> warning C4530: C++ exception handler used, but unwind semantics are
> not enabled. Specify /EHsc C:\Program Files (x86)\Microsoft Visual
> Studio\2017\Community\VC\Tools\MSVC\14.16.27023\include\utility(405):
> error C2440: '<function-style-cast>': cannot convert from 'initializer
> list' to '_Mypair' C:\Program Files (x86)\Microsoft Visual
> Studio\2017\Community\VC\Tools\MSVC\14.16.27023\include\utility(405):
> note: No constructor could take the source type, or constructor
> overload resolution was ambiguous
> ..\examples\json_object\json.cpp(16): note: see reference to function
> template instantiation 'std::pair<std::string,A>
> std::make_pair<std::string,A>(_Ty1 &&,_Ty2 &&)' being compiled with
> [
> _Ty1=std::string,
> _Ty2=A ]
gcc
编译器的类似错误。当我从类中删除std::mutex
或者不在std::vector
上推送对象时,它编译得很好。
作为P.W.指出,std::mutex
既不是可复制的也不是可移动的,并且有充分的理由。拥有互斥锁的重点是防止同时对某些数据进行多线程访问。移动操作本身需要受到保护,移动操作应该使用互斥锁。
以下示例为类提供了一些可移动数据,并显示了如何在移动操作中使用互斥锁(复制操作将类似):
#include <iostream>
#include <mutex>
#include <vector>
#include <memory>
class A {
public:
A() {};
// Move constructor
A(A&& other) {
std::lock_guard<std::mutex> guard(other.m_mutex);
m_data = std::move(other.m_data);
}
// Move operator
A& operator=(A&& other) {
if (this == &other) return *this;
// Lock this and other in a consistent order to prevent deadlock
std::mutex* first;
std::mutex* second;
if (this < &other) {
first = &this->m_mutex;
second = &other.m_mutex;
} else {
first = &other.m_mutex;
second = &this->m_mutex;
}
std::lock_guard<std::mutex> guard1(*first);
std::lock_guard<std::mutex> guard2(*second);
// Now both this and other are safe to access. Do the actual data move.
m_data = std::move(other.m_data);
return *this;
}
private:
std::mutex m_mutex;
std::unique_ptr<int> m_data;
};
int main() {
std::vector<std::pair<std::string,A>> aa;
A a1;
A a2;
a1 = std::move(a2);
aa.emplace_back("aa", std::move(a1));
}
正如P.W所指出的那样,并提供了由怪异提供的暗示,我提出了以下解决方案。
#include <iostream>
#include <mutex>
#include <vector>
#include <memory>
class A {
std::mutex mutex;
public:
A(){};
};
int main()
{
std::vector<std::pair<std::string,std::shared_ptr<A>>> aa;
A a;
//aa.push_back(std::make_pair(std::string("aa"),A()));
//aa.push_back(std::make_pair(std::string("aa"),a));
aa.push_back(std::make_pair(std::string("aa"),std::make_shared<A>()));
}
我修改了我的容器以存储对象的智能指针而不是对象本身。