最近刚从linux C转做android,老大突然看着我闲,叫我去验证一下“一个进程有多个子线程,子线程都注册监听某个信号,另一个进程向它发送该信号的时候,它会怎么处理?”。
带着这个问题,我搜索了各个贴子之后,大概得出:
进程处理信号,你需要注册signal的一个处理函数,线程你需要用signal_wait去等待一个信号。大体得出,如果一个多线程的进程得到了信号,它是会在它诸多子线程里面选一个来执行,有人说是正在进行的那个线程。在多线程环境下,一般会让其他子线程不处理信号,专门用一个线程来处理信号,把异步变成同步处理。
光看人家的贴子是不行的的。为此,我写了如下代码来验证:
#include <stdio.h> #include <signal.h> #include <stddef.h> #include <unistd.h> #include <pthread.h> #include <string.h> static pthread_t g_thread_ids[2]={0}; void ouch1(int sig) { printf("mainthread interrupted,thread id:%u\n",(unsigned int)pthread_self()); //signal(SIGINT,SIG_DFL); } void ouch2(int sig) { printf("child thread 1 interrupted,thread id:%u\n",(unsigned int)pthread_self()); //signal(SIGINT,SIG_DFL); } void ouch3(int sig) { printf("child thread 2 interrupted,thread id:%u\n",(unsigned int)pthread_self()); //signal(SIGINT,SIG_DFL); } void thread_loop1(char* msg) { printf("child thread 1 signal now \n"); signal(SIGINT,ouch2); printf("chilid thread1:%d,%s",(int)getpid(),msg); while(1); } void thread_loop2(char* msg) { printf("child thread 2 signal now\n"); signal(SIGINT,ouch3); printf("child thread2:%d,%s",(int)getpid(),msg); while(1); } void thread_wait2() { //waiting for thread terminate if(g_thread_ids[0]!=0) { pthread_join(g_thread_ids[0],NULL); printf("thread %d terminated\n",getpid()); } if(g_thread_ids[1]!=0) { pthread_join(g_thread_ids[1],NULL); printf("thread %d terminated\n",getpid()); } } //=============test multi-thread test. void start_test() { pthread_t thread1,thread2; char *msg1="this is thread 1\n"; char *msg2="this is thread 2\n"; printf("main thread signal now\n"); signal(SIGINT,ouch1); printf("main thread signal now\n"); pthread_create(&thread1,NULL,(void*)thread_loop1,(void*)msg1); g_thread_ids[0]=thread1; pthread_create(&thread2,NULL,(void*)thread_loop2,(void*)msg2); g_thread_ids[1]=thread2; thread_wait2(); printf("all thread finished its tasks\n"); return ; } int main() { start_test(); return 0; }
说明,跟正在执行的线程没关系,指定一个之后就会一直由它来处理。如果对一个信号注册了多次,那么最后一次有效,其他的都无效。
mark一下。
原文地址:http://blog.csdn.net/minimicall/article/details/39082429