error: ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function
error: ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function
我正在尝试以下代码：
std::thread t(&(Transmitter::sender), this, some_variables);
其中 sender 是同一 class 的成员函数，从其方法调用上面的行。
我收到警告：
Transmitter.h: In member function 'int Transmitter::transmit_streams(std::vector<std::vector<single_stream_record> >, int, Receiver&)':
Transmitter.h:81:44: error: ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function. Say '&Transmitter::sender' [-fpermissive]
尽管它编译并运行良好。我怎样才能删除这个警告。
我的 g++ 是 4.6.3，我用 -std=c++0x 编译代码。
报错信息很清楚。
ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function. Say '&Transmitter::sender' [-fpermissive]
A pointer to member is only formed when an explicit & is used and its operand is a qualified-id not enclosed in parentheses. [ Note: That is, the expression &(qualified-id), where the qualified-id is enclosed in parentheses, does not form an expression of type “pointer to member”. Neither does qualified-id, because there is no implicit conversion from a qualified-id for a non-static member function to the type “pointer to member function” as there is from an lvalue of function type to the type “pointer to function” ([conv.func]). Nor is &unqualified-id a pointer to member, even within the scope of the unqualified-id's class. — end note ]
您需要使用：
std::thread t(&Transmitter::sender, this, some_variables);
我正在尝试以下代码：
std::thread t(&(Transmitter::sender), this, some_variables);
其中 sender 是同一 class 的成员函数，从其方法调用上面的行。
我收到警告：
Transmitter.h: In member function 'int Transmitter::transmit_streams(std::vector<std::vector<single_stream_record> >, int, Receiver&)':
Transmitter.h:81:44: error: ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function. Say '&Transmitter::sender' [-fpermissive]
尽管它编译并运行良好。我怎样才能删除这个警告。
我的 g++ 是 4.6.3，我用 -std=c++0x 编译代码。
报错信息很清楚。
ISO C++ forbids taking the address of an unqualified or parenthesized non-static member function to form a pointer to member function. Say '&Transmitter::sender' [-fpermissive]
A pointer to member is only formed when an explicit & is used and its operand is a qualified-id not enclosed in parentheses. [ Note: That is, the expression &(qualified-id), where the qualified-id is enclosed in parentheses, does not form an expression of type “pointer to member”. Neither does qualified-id, because there is no implicit conversion from a qualified-id for a non-static member function to the type “pointer to member function” as there is from an lvalue of function type to the type “pointer to function” ([conv.func]). Nor is &unqualified-id a pointer to member, even within the scope of the unqualified-id's class. — end note ]
您需要使用：
std::thread t(&Transmitter::sender, this, some_variables);