我开始在 Linux 上进行 pthread 编程,在第一个程序中我完全感到困惑。下面是我正在运行的程序
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
void *print_message_function( void *ptr );
int main(){
pthread_t thread1, thread2;
char *message1 = "Thread 1";
char *message2 = "Thread 2";
int iret1, iret2;
/* Create independent threads each of which will execute function */
iret1 = pthread_create( &thread1, NULL, print_message_function, (void*) message1);
iret2 = pthread_create( &thread2, NULL, print_message_function, (void*) message2);
/* Wait till threads are complete before main continues. Unless we */
/* wait we run the risk of executing an exit which will terminate */
/* the process and all threads before the threads have completed. */
pthread_join( thread1, NULL);
printf("amit");
pthread_join( thread2, NULL);
printf("Thread 1 returns: %d\n",iret1);
printf("Thread 2 returns: %d\n",iret2);
exit(0);
}
void *print_message_function( void *ptr ){
char *message;
message = (char *) ptr;
printf("%s \n", message);
}
我想知道的第一件事是线程执行的顺序不是连续的?
第二件事是我故意把 print("amit");可以看到 main 在 thread1 终止期间确实停止了,但在输出中我们可以看到 printf 语句首先被执行。整个过程的输出为
Thread 1
Thread 2
amitThread 1 返回:0
线程 2 返回:0
你说得对线程执行顺序不连续。在某种程度上,这就是使用线程的全部意义,即同时运行其他任务。
您看到的输出与预期一致,并且可能有所不同。
也许这会有所帮助:
main thread1 thread2
|
|--create--------+-----------\
| | |
| "Thread 1" | "Thread 2" can
| | |<- occur anywhere
| / | along this line
join(1) --------- |
| |
| |
"amit" |
| |
| |
join(2) ---------------------/
|
|
"Thread 1 returns"
"Thread 2 returns"
|
exit(0)
您唯一的保证是:
- "
Thread 1
“总是会在之前打印”amit
“ (因为pthread_join()
等待线程 1 结束,然后主程序才能继续)
- "
Thread X returns ...
" 语句总是出现在最后,两个线程都终止之后。
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