Question

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;
}

Answer 1

如果没有可用的数据，在经过5秒超时后，您的选择呼叫将返回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。

为什么以下打印'资源暂时不可用'？

1 个答案: