Why does the following code print ‘read(): Resource temporarily unavailable’ 80% of the time? That is the EAGAIN code, which is the same as WOULD BLOCK which means there is no data waiting to be read, but select is returning 1 saying there is data (tested in Linux):
为什么以下代码打印'read():资源暂时不可用'80%的时间?这是EAGAIN代码,它与WOULD BLOCK相同,这意味着没有数据等待读取,但是select返回1表示有数据(在Linux中测试):
#include <time.h>
#include <unistd.h>
#include <stdio.h>
#include <string.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <sys/errno.h>
int main(int argc, char** argv)
{
int fd = open("/dev/lp0", O_RDWR | O_NONBLOCK);
int ret = 0;
int status = 0;
char buffer[1024];
char teststr[] = "This is a test\n";
char XMIT_STATUS_OFFLINE[] = {0x10,0x04,0x02};
char XMIT_STATUS_ERROR[] = {0x10,0x04,0x03};
char XMIT_STATUS_ROLL[] = {0x10,0x04,0x04};
char XMIT_STATUS_SLIP[] = {0x10,0x04,0x05};
fd_set rfds;
FD_ZERO( &rfds );
FD_SET( fd, &rfds );
struct timeval sleep;
sleep.tv_sec = 5;
sleep.tv_usec = 0;
/* Offline status */
ret = write(fd, XMIT_STATUS_OFFLINE, sizeof(XMIT_STATUS_OFFLINE));
//printf("write() returned %d\n", ret);
do {
ret = select( fd + 1, &rfds, NULL, NULL, &sleep );
} while (ret < 0 && (errno == EINTR));
ret = read(fd, buffer, 1024);
if(ret == -1) {
perror("read(): ");
} else {
status = buffer[0];
if((status & 0x04) != 0)
{
printf("The cover is open.\n");
} else {
printf("OFFLINE is good.\n");
}
}
close(fd);
return 0;
}
1 个解决方案
#1
6
Your select call will return 0 after the 5 second timeout elapses if no data is available. Your code will ignore this and try to read from the device anyways. Check for ret == 0 and that will fix your problem.
如果没有可用数据,则在5秒超时后,您的选择调用将返回0。您的代码将忽略此操作并尝试从设备读取。检查ret == 0,这将解决您的问题。
#1
6
Your select call will return 0 after the 5 second timeout elapses if no data is available. Your code will ignore this and try to read from the device anyways. Check for ret == 0 and that will fix your problem.
如果没有可用数据,则在5秒超时后,您的选择调用将返回0。您的代码将忽略此操作并尝试从设备读取。检查ret == 0,这将解决您的问题。