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
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
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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