对于参数 ‘sender’ 到 ‘void*’，C++ 无法将 ‘1’ 转换为 ‘void* sending(void*)’

Question

我在 sender 类中有 startSending 过程和一个友元函数 (sending)。我想从一个新线程调用 friend 函数，所以我在 startSending 过程中创建了一个新线程。

class sender{
    public:
    void startSending();

    friend void* sending (void * callerobj);
}

void sender::startSending(){
    pthread_t tSending;
    pthread_create(&tSending, NULL, sending(*this), NULL);
    pthread_join(tSending, NULL);
}

void* sending (void * callerobj){
}

但是我得到了这个错误

cannot convert ‘sender’ to ‘void*’ for argument ‘1’ to ‘void* sending(void*)’

从 pthread_create 调用发送的正确方法是什么？

Answer 1

pthread_create 签名如下所示:

int pthread_create(pthread_t *thread, //irrelevant
                   const pthread_attr_t *attr, //irrelevant
                   void *(*start_routine) (void *), //the pointer to the function that is going to be called
                   void *arg); //the sole argument that will be passed to that function

所以在你的情况下，指向发送的指针必须作为第三个参数传递，而this(将传递的参数to sending) 需要作为最后一个参数传递:

pthread_create(&tSending, NULL, &sending, this);