#include <sys/eventfd.h>
#include <sys/timerfd.h>
#include <unistd.h>
#include <stdio.h>
#include <stdint.h>
#include <stdlib.h>
#include <errno.h>
#include <pthread.h>
int evfd;
void * f(void *p)
{
int ret = 0,j = 100;
while(ret >= 0)
{
uint64_t i = 0;
fd_set readfds,allfds;
FD_SET(evfd,&readfds);
FD_SET(evfd,&allfds);
int fds = select(evfd + 1,&readfds,0,0,0);
if(FD_ISSET(evfd,&allfds))
{
ret = read(evfd,&i,sizeof(int64_t));
printf("%d,%d,%llu\n",pthread_self(),ret,i);
}
}
printf("pid=%d exit\n",pthread_self());
}
int main(int argc, char *argv[])
{
//evfd = eventfd(1000,EFD_NONBLOCK|EFD_CLOEXEC);
evfd = eventfd(1000,0);
pthread_t pid1 = 0,pid2 = 0,pid3 = 0;
pthread_create(&pid1,0,f,0);
//pthread_create(&pid2,0,f,0);
//pthread_create(&pid3,0,f,0);
printf("pid=%d,%d,%d\n",pid1,pid2,pid3);
sleep(1);
uint64_t j = 100;
int ret = write(evfd,&j,sizeof(uint64_t));
printf("write j=%d,%d\n",j,ret);
sleep(1);
j++;
ret = write(evfd,&j,sizeof(uint64_t));
printf("write j=%d,%d\n",j,ret);
//这句注释,得出的结论差异很大.
//应该多次(连续两次)写,读方一次读,8 字节的内核计数缓冲区写乱了
//所以,证明write 能触发读操作,但是具体触发几次读操作,还需要读
//操作自己控制,否则写次数 != 读次数,理论上要完全匹配也很难
//操作上应该是触发读后,读操作后的动作需要合理的判断结束点
//sleep(1);
j++;j++;
ret = write(evfd,&j,sizeof(uint64_t));
printf("write j=%d,%d\n",j,ret);
sleep(1);
j++;j++;
ret = write(evfd,&j,sizeof(uint64_t));
printf("write j=%d,%d\n",j,ret);
sleep(1);
close(evfd);
return 0;
}
输出：
gcc cc.c -o cc -lpthread;./cc
pid=-1553656064,0,0
-1553656064,8,1000
write j=100,8
-1553656064,8,100
write j=101,8
write j=103,8
-1553656064,8,204
write j=105,8
-1553656064,8,105
这里，连续写了两次，读了一次，导致缓冲区乱了。
write j=101,8
write j=103,8
-1553656064,8,204