C ++ 11线程初始化带有成员函数编译错误 [英] C++ 11 Thread initialization with member functions compiling error

问题描述

我刚刚开始使用C ++ 11线程，我一直在挣扎（可能是愚蠢的）错误。
这是我的示例程序：

  #include< iostream& 
 #include< thread> 
 #include< future> 
 using namespace std; 
 
 class A {
 public：
 A（）{
 cout< A constructor\\\
; 
} 
 
 void foo（）{
 cout< 我是foo（），我问候你。 
} 
 
 static void foo2（）{
 cout< 我是foo2（），我是静态的！\\\
; 
} 
 
 void operator（）（）{
 cout< 我是运算符（）。Hi there！\\\
; 
} 
}; 
 
 void hello1（）{
 cout< 你从外面A类 
} 
 
 int main（）{
 A obj; 
 thread t1（hello1）; // it works 
 thread t2（A :: foo2）; // it works 
 thread t3（obj.foo）; // error 
 thread t4（obj）; // it works 
 
 t1.join（）; 
 t2.join（）; 
 t3.join（）; 
 t4.join（）; 
 return 0; 
} 
  

是否可以从纯成员函数启动线程？如果不是，我如何从对象 obj 包装我的 foo 函数，以便能够创建这样的线程？
提前感谢！

这是编译错误：

thread_test.cpp：在函数'int main（）'：
thread_test.cpp：32：22：error：没有匹配的函数调用'std :: thread :: thread（）'

thread_test.cpp：32：22：注意：候选人是：

/ usr / include / c ++ / 4.6 / thread：133 ：7：注意：std :: thread :: thread（_Callable&&&&&& ...）[with _Callable = void（A :: *）（），_Args = {}]

/ usr / include / c ++ / 4.6 / thread：133：7：注意：没有已知的转换从参数1从''到'void（A :: *&& ）'

/ usr / include / c ++ / 4.6 / thread：128：5：note：std :: thread :: thread（std :: thread&&

/ usr / include / c ++ / 4.6 / thread：128：5：注意：参数1从''到'std :: thread&& '

/ usr / include / c ++ / 4.6 / thread：124：5：note：std :: thread :: thread（）

/ usr / include / c ++ / 4.6 / thread：124：5：注意：候选者期望有0个参数，提供1个

解决方案

您需要一个不带参数的可调用对象，因此

  （& A :: foo，& obj）; 
  

这具有创建可调用实体的效果，该实体在 obj 上调用 A :: foo 。

原因是 A 的非静态成员函数采用类型的隐式第一个参数（可能是cv限定） A * 。当你调用 obj.foo（）时，你正在有效地调用 A :: foo（& obj）。一旦你知道，上面的咒语是完全有道理的。

I'm just starting to use C++ 11 threads and I've been struggling on a (probably silly) error. This is my example program:

#include <iostream>
#include <thread>
#include <future>
using namespace std;

class A {
public:
  A() {
    cout << "A constructor\n";
  }

  void foo() {
    cout << "I'm foo() and I greet you.\n";
  }

  static void foo2() {
    cout << "I'm foo2() and I am static!\n";
  }

  void operator()() {
    cout << "I'm the operator(). Hi there!\n";
  }
};

void hello1() {
  cout << "Hello from outside class A\n";
}

int main() {
  A obj;
  thread t1(hello1); //  it works
  thread t2(A::foo2); // it works
  thread t3(obj.foo); // error
  thread t4(obj);     // it works

  t1.join();
  t2.join();
  t3.join();
  t4.join();
  return 0;
}

Is it possible to start a thread from a pure member function? If it is not, how can I wrap my foo function from object obj to be able to create such thread? Thanks in advance!

This is the compiling error:

thread_test.cpp: In function ‘int main()’: thread_test.cpp:32:22: error: no matching function for call to ‘std::thread::thread()’

thread_test.cpp:32:22: note: candidates are:

/usr/include/c++/4.6/thread:133:7: note: std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (A::*)(), _Args = {}]

/usr/include/c++/4.6/thread:133:7: note: no known conversion for argument 1 from ‘’ to ‘void (A::*&&)()’

/usr/include/c++/4.6/thread:128:5: note: std::thread::thread(std::thread&&)

/usr/include/c++/4.6/thread:128:5: note: no known conversion for argument 1 from ‘’ to ‘std::thread&&’

/usr/include/c++/4.6/thread:124:5: note: std::thread::thread()

/usr/include/c++/4.6/thread:124:5: note: candidate expects 0 arguments, 1 provided

解决方案

You need a callable object taking no parameters, so

thread t3(&A::foo, &obj);

should do the trick. This has the effect of creating a callable entity which calls A::foo on obj.

The reason is that a non-static member function of A takes an implicit first parameter of type (possibly cv qualified) A*. When you call obj.foo() you are effectively calling A::foo(&obj). Once you know that, the above incantation makes perfect sense.

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