为什么以下打印“资源暂时不可用”?
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为什么下面的代码80%的时间会打印“read():resource temporary unavailable”?这是eagain代码,与block相同,这意味着没有等待读取的数据,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;
}


最佳答案:

如果没有可用的数据,您的select调用将在5秒超时后返回0。您的代码将忽略此问题,并尝试从设备中读取。检查ret==0,这将解决您的问题。

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