为什么 pthread_join() 的第二个参数是一个 **，一个指向指针的指针？

Question

我是 pthread 的新手，也不熟悉指向指针的指针。有人可以解释为什么 pthread_join() 的第二个参数是 void **。为什么要这样设计。

int pthread_join(pthread_t thread, void **value_ptr);

Answer 1

要通过函数的参数 return 一个值，您需要传入变量的地址以接收新值。

因为 pthread_join() is designed to receive the pointer value being passed to pthread_exit(), which is a void*, pthread_join() 期望 void* 的地址，实际上是 void**.

类型

示例：

#include <stdlib.h> /* for EXIT_xxx macros */
#include <stdio.h> /* for printf() and perror() */
#include <pthread.h> 

void * tf(void * pv)
{
  int * a = pv;
  size_t index = 0;

  printf("tf(): a[%zu] = %d\n", index , a[index]);

  ++index;

  pthread_exit(a + index); /* Return from tf() the address of a's 2nd element. 
                          a + 1 here is equivalent to &a[1]. */
}


int main(void)
{
  int a[2] = {42, 43};
  pthread_t pt;
  int result = pthread_create(&pt, NULL, tf, a); /* Pass to tf() the address of 
                                                    a's 1st element. a decays to 
                                                    something equivalent to &a[0]. */
  if (0 != result)
  {
    perror("pthread_create() failed");
    exit(EXIT_FAILURE);
  }

  {
    int * pi;
    size_t index = 0;

    {
      void * pv;
      result = pthread_join(pt, &pv); /* Pass in the address of a pointer-variable 
                                         pointing to where the value passed to 
                                         pthread_exit() should be written. */
      if (0 != result) 
      {
        perror("pthread_join() failed");
        exit(EXIT_FAILURE);
      }

      pi = pv;
    }

    ++index;

    printf("main(): a[%zu] = %d\n", index, pi[0]);
  }

  return EXIT_SUCCESS;
}

上面的程序预计打印：

tf(): a[0] = 42
main(): a[1] = 43

为什么 pthread_join() 的第二个参数是一个 **，一个指向指针的指针？

Why the second argument to pthread_join() is a **, a pointer to a pointer?

c

pointers

pthreads

pass-by-value

pointer-to-pointer