从输入中读取C/UNIX(受字符数和超时限制) [英] C/UNIX read from input (limited by char count and timeout)

问题描述

在学习决赛时，我发现了一个非常有趣的问题.这就是我想要编写的代码.

while studying for my finals I found a very interesting questing. This is what I desire to code.

程序将stdin读入缓冲区(具有固定大小).当缓冲区已满时，程序会将其打印到文件中.但是，如果缓冲区没有在固定的时间(超时)内被填充，程序将打印到文件[TIMEOUT]，其余的缓冲区将被打印(当前已读取)

Program read stdin into buffer (of fixed size). When buffer is full, program prints it to file. But if buffer isn't filled in fixed time (timeout), program prints to file [TIMEOUT] and the rest of buffer (currently read)

第一个示例:

buffer_size = 5;超时= 4;

buffer_size = 5; timeout = 4;

$ while : ; do printf 1; sleep 1; done | ./a.out

应该写[TIMEOUT]1111[TIMEOUT]1111[TIMEOUT]1111等，因为while循环仅写4个字符(在4秒内).

should write [TIMEOUT]1111[TIMEOUT]1111[TIMEOUT]1111 etc. because while-loop writes only 4 chars (within 4 second limit).

第二个示例

buffer_size = 3;超时= 5;

buffer_size = 3; timeout = 5;

$ while : ; do printf 1; sleep 1; done | ./a.out

应该写111 111 111等，因为while循环写了3个字符(在3秒内<5秒以内)，所以永远不会发生超时.

should write 111 111 111 etc. because while-loop writes 3 chars (within 3 seconds < 5 sec limit) so timeout never happens.

我正在尝试使用poll对其进行编码，但是我不知道如何找出所有字符是否已写入或仅一个字符.我也不能卡在read(0, buffer, buffer_size)上，因为我会错过超时.可能吗?我想这是我们的老师指出的很好的练习.

I'm trying to code it using poll but I don't know how to find out, whether all chars have been written or only one. I can't also get stuck on read(0, buffer, buffer_size) as I would miss timeout. Is it even possible? I guess it is as our teacher pointed it out as a good excersice.

当然，繁忙的等待是不可接受的，只允许使用经典的POSIX系统调用(轮询，选择，读取，写入，打开...).

Of course, busy wait is unacceptable, only classic POSIX syscalls allowed (poll, select, read, write, open...).

有人可以提示我吗?我不知道如何管理这种行为，并且堆栈溢出过多，谷歌也没有给我答案(或者也许我只是不知道要搜索什么)

Could anybody hint me, please? I have no clue how to manage this behaviour and neighter stackoverflow nor google gave me answer (or maybe I just don't know what to search for)

预先感谢

推荐答案

谢谢大家，我以其他方式解决了这个问题(但仍然没有忙碌的等待时间).我附上下面的代码供更多的学生使用:)

thanks guys, I figured it out in other way (but still no busy-wait). I attach the code below for further students :)

#include <stdio.h>
#include <unistd.h>
#include <string.h>
#include <strings.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <libgen.h>
#include <err.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <arpa/inet.h>
#include <time.h>
#include <fcntl.h>
#include <sys/ioctl.h>
#include <sys/poll.h>
#include <sys/time.h>
#include <netinet/in.h>
#include <errno.h>

extern char ** environ;

/*
usage:

$ while : ; do printf 1; sleep 1; done | XTIMEOUT=4 XSIZE=5 ./a.out
[TIMEOUT]1111[TIMEOUT]1111[TIMEOUT]1111...

$ while : ; do printf 1; sleep 1; done | XTIMEOUT=5 XSIZE=3 ./a.out
111...
*/

uint64_t
now()
{
    struct timeval tv;
    gettimeofday(&tv, NULL);
    double time_in_mill =  (tv.tv_sec) * 1000 + (tv.tv_usec) / 1000;
    return (uint64_t) time_in_mill;
}

int
main(int argc, char ** argv) { 

    //  ----------------------------------
    //  boring stuff

    size_t timeout = 11 * 1000;
    size_t buffer_size = 10;

    char * tmp_env;
    if ((tmp_env = getenv("XTIMEOUT")) != NULL) {
        timeout = atoi(tmp_env) * 1000;
    }
    if ((tmp_env = getenv("XSIZE")) != NULL) {
        buffer_size = atoi(tmp_env);
    }

    //  ----------------------------------
    //  fun starts here

    //  buffers
    char * buffer = (char *) malloc(buffer_size * sizeof(char));
    char * buffer2 = (char *) malloc(buffer_size * sizeof(char));

    //  set stdin non-blocking
    int saved_flags = fcntl(0, F_GETFL);
    fcntl(0, saved_flags | O_NONBLOCK);


    //  poll structure
    struct pollfd ufds[1];
    ufds[0].fd = 0;
    ufds[0].events = POLLIN;

    int rv, n, k;
    size_t pos = 0;
    uint64_t start_time;

    int rem_time = timeout;

    for (;;) {
        start_time = now();
        //printf("pollig for %d\n", rem_time);
        rv = poll(ufds, 1, rem_time);

        if (rv == -1) {     //  err
            err(1, "poll");
        }
        else if (rv == 0) { //  timeout
            write(1, "[TIMEOUT]", 9);
            write(1, buffer, pos);
            pos = 0;
            rem_time = timeout;
        }
        else {      //  regular
            if (ufds[0].revents & POLLIN) {     //  real action
                if ((n = read(0, buffer2, buffer_size-pos)) == -1) {    //  read up to free space
                    err(1, "read");
                }

                for (k = 0; k < n; ++k) {   //  copy (maybe strcp)
                    buffer[k+pos] = buffer2[k];
                }

                pos += n;   //  pos has changed

                if (pos == buffer_size) {   //  buffer is full -> write it out and set new timeout
                    write(1, buffer, buffer_size); write(1, "", 1);
                    pos = 0;
                    rem_time = timeout;
                }
                else {      // still not enough, make timeout shorter (by the length of computation)
                    rem_time = rem_time - now() + start_time;
                }
            }
            else {      //  poll failed!
                err(1, "false alarm");
            }
        }
    }

    free(buffer);
    free(buffer2);

    return (0);
}

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