One solution is to force mainto wait until the other two threads are done.What we
need is a function similar to waitthat waits for a thread to finish instead of a process.
That function is pthread_join,which takes two arguments: the thread ID of the
thread to wait for, and a pointer to a void*variable that will receive the finished
thread’s return value. If you don’t care about the thread return value, pass NULLas the
second argument.
一个解决办法是强迫main函数等待另外两个线程的结束。我们需要一个类似wait的函
数，但是等待的是线程而不是进程。这个函数是pthread_join。它接受两个参数：线程ID，
和一个指向void*类型变量的指针，用于接收线程的返回值。如果你对线程的返回值不感兴
趣，则将NULL作为第二个参数。
The moral of the story: Make sure that any data you pass to a thread by reference is
not deallocated,even by a different thread, until you’re sure that the thread is done with
it.This is true both for local variables, which are deallocated when they go out of
scope, and for heap-allocated variables, which you deallocate by calling free(or using
deletein C++).
int main ()
{
pthread_t thread1_id;
pthread_t thread2_id;
struct char_print_parms thread1_args;
struct char_print_parms thread2_args;
/* Create a new thread to print 30,000 x’s. */
thread1_args.character = ’x’;
thread1_args.count = 30000;
pthread_create (&thread1_id, NULL, &char_print, &thread1_args);
/* Create a new thread to print 20,000 o’s. */
thread2_args.character = ’o’;
thread2_args.count = 20000;
pthread_create (&thread2_id, NULL, &char_print, &thread2_args);
/* Make sure the first thread has finished. */
pthread_join (thread1_id, NULL);
/* Make sure the second thread has finished. */
pthread_join (thread2_id, NULL);
/* Now we can safely return. */
return 0;
}