无法将对抽象 class 的引用传递给线程函数?

Reference to abstract class can't be passed to thread function?

我想在使用 std::thread 调用的函数中使用对抽象 class (A) 的引用作为参数类型。 这似乎是不可能的,因为编译器出于某种原因尝试编译:std::tuple<A>,即使在我的代码中只有 A 的引用类型被用作参数(从不作为值类型)。

#include <cstdlib>
#include <thread>

class A {
public:
  virtual void a() = 0;
};

class B : public A {
public:
  virtual void a() {
  }
};

class C {
public:
  C(A& aRef) {
    thread = std::thread(&C::doSomething, this, aRef);
  }
private:
  void doSomething(A& aRef) {

  }
  std::thread thread;
};

int main(int argc, char* argv[]) {
  B b;
  C c(b);
  return EXIT_SUCCESS;
}

将在 Visual Studio 2017 年输出:

error C2259: 'A': cannot instantiate abstract class
tuple(278): note: see reference to class template instantiation 'std::tuple<A>' being compiled
tuple(278): note: see reference to class template instantiation 'std::tuple<C *,A>' being compiled
thread(49): note: see reference to class template instantiation 'std::tuple<void (__thiscall C::* )(A &),C *,A>' being compiled
main.cpp(18): note: see reference to function template instantiation 'std::thread::thread<void(__thiscall C::* )(A &),C*const ,A&,void>(_Fn &&,C *const &&,A &)' being compiled

为什么 std::thread 尝试编译 std::tuple<A>?如果我直接从主线程调用 C::doSomething,代码编译正常。

这里有什么我遗漏的吗?

当您将引用作为参数传递给您的线程时,您需要将引用包装在 std::reference_wrapper 中。

您可以使用 std::ref() 来做到这一点。

看看这个例子:

#include <thread>
#include <utility>
#include <iostream>

struct A {
    virtual void a() = 0;
};

struct B : public A {
    virtual void a() { std::cout << "Hello World\n"; }
};

class C {
    void doSomething(A& aRef) { aRef.a(); }
    std::thread thread;
public:
    C(A& aRef) : thread(&C::doSomething, this, std::ref(aRef)) {}
    // or alternatively using a lambda:
    //C(A& aRef) : thread( [&](){ doSomething(aRef); } ) {}
    void join() { thread.join(); }
};

int main(int argc, char* argv[]) {
    B b;
    C c(b);
    c.join();
}

这样编译运行:

$ g++ ~/tt.cc -std=c++11 -pthread && ./a.out
Hello World

供参考: