I am new to using pthread and also not that familiar with pointers to pointers. Could somebody perhaps explain why the second argument of pthread_join() is a void **. Why is it designed like this.
int pthread_join(pthread_t thread, void **value_ptr);
To return a value via a function's argument you need to pass in the address of the variable to receive the new value.
As pthread_join() is designed to receive the pointer value being passed to pthread_exit(), which is a void*, pthread_join() expects the address of a void* which in fact is of type void**.
Example:
#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;
}
The program above is expected to print:
tf(): a[0] = 42
main(): a[1] = 43
