pthread_create不起作用.传递参数3警告

Question

我正在尝试创建一个线程,从我记得这应该是正确的方法:

#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#define NUM_THREADS 5

int SharedVariable =0;
void SimpleThread(int which)
{
    int num,val;
    for(num=0; num<20; num++){
        if(random() > RAND_MAX / 2)
            usleep(10);
        val = SharedVariable;
        printf("*** thread %d sees value %d\n", which, val);
        SharedVariable = val+1;
    }
    val=SharedVariable;
    printf("Thread %d sees final value %d\n", which, val);
}

int main (int argc, char *argv[])
{
   pthread_t threads[NUM_THREADS];
   int rc;
   long t;
   for(t=0; t< NUM_THREADS; t++){
      printf("In main: creating thread %ld\n", t);
      rc = pthread_create(&threads[t], NULL, SimpleThread, (void* )t);
      if (rc){
         printf("ERROR; return code from pthread_create() is %d\n", rc);
         exit(-1);
      }
   }

   /* Last thing that main() should do */
   pthread_exit(NULL);
}

而我得到的错误是这个:

test.c:在函数'main'中:test.c:28:警告:从不兼容的指针类型/usr/include/pthread.h:227传递'pthread_create'的参数3:注意:expected'void*(*)( void*)'但参数的类型为'void(*)(int)'

我无法更改SimpleThread函数,因此更改参数的类型不是一个选项,即使我已经尝试过它也不起作用.

我究竟做错了什么？

Answer 1

SimpleThread 应该声明为

void* SimpleThread(void *args) {
}

当您将参数传递给你的线程,最好是定义一个struct对他们来说,一个指针传递给struct作为void*,并投退给函数内部的权利类型.