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 *)’
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void *。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。