无法正确创建线程

时间:2014-01-28 21:59:38

标签: c++ c multithreading pthreads

有人可以告诉我,我做错了什么吗? (我省略了程序的其余部分,因为它很长......)

#include <pthread.h>

void *RTPfun(char *client_addr);

int main(int argc, char *argv[])
{ 
  char* client_addr;
  pthread_t RTPthread;

  // ...

  pthread_create(&RTPthread, NULL, &RTPfun, client_addr) 
}

void *RTPfun(char * client_addr)
{
  // ...
  return;
}

错误:

TCPserver.c: In function ‘main’:
TCPserver.c:74:5: warning: passing argument 3 of ‘pthread_create’ from incompatible pointer type [enabled by default]
/usr/include/pthread.h:225:12: note: expected ‘void * (*)(void *)’ but argument is of type ‘void * (*)(char *)’

2 个答案:

答案 0 :(得分:2)

Pthread使用接收void *并返回void *。

的函数

您需要将函数的参数从char *更改为v​​oid *。这是另一种选择:

#include <pthread.h>



void *RTPfun(void *client_addr);


int main(int argc, char *argv[])
{ 
  char* client_addr;
  pthread_t RTPthread;

   ...
   ...

  pthread_create(&RTPthread, NULL, &RTPfun, client_addr) 
}



void *RTPfun(void* data)
{
 char *client_addr = (char*)data;
 ....
 return;
}

答案 1 :(得分:1)

您必须将char指针转换为void指针。

#include <pthread.h>

void *RTPfun(void *client_addr);

int main(int argc, char *argv[])
{ 
  char* client_addr;
  pthread_t RTPthread;

   ...
   ...

  pthread_create(&RTPthread, NULL, &RTPfun, (void*)client_addr) 
}

void *RTPfun(void * client_addr)
{
 char *something = (char*)client_addr;
 ....
 return;
}

每次需要传递一些数据时都会使用Void指针,而事先无法知道变量的类型(char *,integer * ...)。您赋予pthread_create的函数将void *作为输入,因此您可以将char指针强制转换为void函数,并在RTPfun中执行相反的操作。