为什么 pthread_create 的 start_routine 返回 void* 并取 void*

Question

The function header for pthread_create looks like this:

int pthread_create(pthread_t * thread, 
                   const pthread_attr_t * attr,
                   void * (*start_routine)(void *), 
                   void *arg);

I understand it all except that the function pointer for start_routine is of the form void* (*fpointer) (void*) which means it takes in a void* and returns a void* .

The void* parameter is just a way to pass in an argument to the start_routine , I get that part, but I don't understand why the function returns a void* ? What code will even notice that return value?

Answer 1

From the documentation for pthread_create :

The thread is created executing start_routine with arg as its sole argument. If the start_routine returns, the effect is as if there was an implicit call to pthread_exit() using the return value of start_routine as the exit status. Note that the thread in which main() was originally invoked differs from this. When it returns from main() , the effect is as if there was an implicit call to exit() using the return value of main() as the exit status.

And pthread_exit :

The pthread_exit() function terminates the calling thread and makes the value value_ptr available to any successful join with the terminating thread.

So if you do a pthread_join on a thread, the pointer it returns is passed back to the joining thread, allowing you to transmit information from the dying thread to another, living thread.

Answer 2

From the spec :

If the start_routine returns, the effect is as if there was an implicit call to pthread_exit() using the return value of start_routine as the exit status.