重定向运算符在 unix 中不起作用？

Question

我正在运行开发一个程序，它正在控制台上打印所有 printf 语句。但是当我尝试使用“>”将它们重定向到任何文件时，该文件已创建但文件中没有程序输出。请帮忙当我运行控制台中的以下代码时：

#include <stdio.h>
#include <math.h>
#include <sys/time.h>
#include <pthread.h>
#include <stdlib.h>
#include <unistd.h>
#include <ctype.h>

double time_diff(struct timeval x , struct timeval y);

struct timeval initial;
long sno=0;

void *process1 (void *sleepTimeForP1);
void *process2 (void *sleepTimeForP2);

pthread_mutex_t lock;

int main()
{
    gettimeofday(&initial , NULL);
    pthread_t trd1,trd2;
    int thread1,thread2;
    int var1;
    int var2;
    int *sleepTimeForP1;
    int *sleepTimeForP2;
    var1=rand()%9+1;
    sleepTimeForP1=&var1;
    var2=rand()%9+1;
    sleepTimeForP2=&var2;
    printf("S No.\tThread Number\tItem\tTime(usec)\n");
    thread1=pthread_create(&trd1,NULL,process1,(void *)sleepTimeForP1);
    thread2=pthread_create(&trd2,NULL,process2,(void *)sleepTimeForP2);
    pthread_join(trd1, NULL);
    pthread_join(trd2, NULL);
    printf("pthread1 = %d\n",thread1);
    printf("pthread2 = %d\n",thread2);

    return 0;
}

double time_diff(struct timeval x , struct timeval y)
{
    double x_ms , y_ms , diff;
    x_ms = (double)x.tv_sec*1000000 + (double)x.tv_usec;
    y_ms = (double)y.tv_sec*1000000 + (double)y.tv_usec;
    diff = (double)y_ms - (double)x_ms;
    return diff;
}

void *process1 (void *sleepTimeForP1)
{
    int *tsleepTimeForP1 = (int *)sleepTimeForP1;
    struct timeval end;
    while(1)
    {
        pthread_mutex_lock(&lock);
        sno++;
        gettimeofday(&end , NULL);
        printf("%ld\t1\t\t1.1\t%.0lf\n",sno,time_diff(initial, end));
        sno++;
        gettimeofday(&end , NULL);
        printf("%ld\t1\t\t1.2\t%.0lf\n",sno,time_diff(initial, end));
        pthread_mutex_unlock(&lock);
        sleep(1);
    }

}

void *process2 (void *sleepTimeForP2)
{
    int *tsleepTimeForP2 = (int *)sleepTimeForP2;
    struct timeval end;
    while(1)
    {
        pthread_mutex_lock(&lock);
        sno++;
        gettimeofday(&end , NULL);
        printf("%ld\t2\t\t2.1\t%.0lf\n",sno,time_diff(initial, end));
        sno++;
        gettimeofday(&end , NULL);
        printf("%ld\t2\t\t2.2\t%.0lf\n",sno,time_diff(initial, end));
        pthread_mutex_unlock(&lock);
        sleep(1);
    }
}

它给我以下输出：

S No.   Thread Number   Item    Time(usec)
1   1       1.1 320
2   1       1.2 438
3   2       2.1 506
4   2       2.2 586
5   1       1.1 1000592
6   1       1.2 1000629
7   2       2.1 1000714
8   2       2.2 1000740
9   1       1.1 2000820
10  1       1.2 2000927
11  2       2.1 2000998
12  2       2.2 2001099
13  1       1.1 3001165
14  1       1.2 3001285
15  2       2.1 3001355
16  2       2.2 3001441
17  1       1.1 4001518
18  1       1.2 4001635
19  2       2.1 4001706
20  2       2.2 4001798
21  1       1.1 5001776

但是当我这样做时 ./a.out > b.txt 我在控制台和文件中都没有得到任何输出

Answer 1

我会给予一些指点。阅读 Linux Programming Book，了解实际情况。

我在你的代码的相关 printf() 语句后添加了一个 fflush(stdout)，即 line 32,65,85.

完成此操作后，您将获得所需的输出。现在尝试理解为什么会出现这种行为以及为什么需要明确的 fflush()？我相信问题 poster 应该做出一些努力来理解问题并 post a Minimal Complete Verifiable Example

完整代码：

#include <stdio.h>
#include <math.h>
#include <sys/time.h>
#include <pthread.h>
#include <stdlib.h>
#include <unistd.h>
#include <ctype.h>

double time_diff(struct timeval x , struct timeval y);

struct timeval initial;
long sno=0;

void *process1 (void *sleepTimeForP1);
void *process2 (void *sleepTimeForP2);

pthread_mutex_t lock;

int main()
{
    gettimeofday(&initial , NULL);
    pthread_t trd1,trd2;
    int thread1,thread2;
    int var1;
    int var2;
    int *sleepTimeForP1;
    int *sleepTimeForP2;
    var1=rand()%9+1;
    sleepTimeForP1=&var1;
    var2=rand()%9+1;
    sleepTimeForP2=&var2;
    printf("S No...\tThread Number\tItem\tTime(usec)\n");
    fflush(stdout);
    thread1=pthread_create(&trd1,NULL,process1,(void *)sleepTimeForP1);
    thread2=pthread_create(&trd2,NULL,process2,(void *)sleepTimeForP2);
    pthread_join(trd1, NULL);
    pthread_join(trd2, NULL);
    printf("pthread1 = %d\n",thread1);
    printf("pthread2 = %d\n",thread2);

    return 0;
}

double time_diff(struct timeval x , struct timeval y)
{
    double x_ms , y_ms , diff;
    x_ms = (double)x.tv_sec*1000000 + (double)x.tv_usec;
    y_ms = (double)y.tv_sec*1000000 + (double)y.tv_usec;
    diff = (double)y_ms - (double)x_ms;
    return diff;
}

void *process1 (void *sleepTimeForP1)
{
    int *tsleepTimeForP1 = (int *)sleepTimeForP1;
    struct timeval end;
    while(1)
    {
        pthread_mutex_lock(&lock);
        sno++;
        gettimeofday(&end , NULL);
        printf("%ld\t1\t\t1.1\t%.0lf\n",sno,time_diff(initial, end));
        sno++;
        gettimeofday(&end , NULL);
        printf("%ld\t1\t\t1.2\t%.0lf\n",sno,time_diff(initial, end));
        fflush(stdout);
        pthread_mutex_unlock(&lock);
        sleep(1);
    }

}

void *process2 (void *sleepTimeForP2)
{
    int *tsleepTimeForP2 = (int *)sleepTimeForP2;
    struct timeval end;
    while(1)
    {
        pthread_mutex_lock(&lock);
        sno++;
        gettimeofday(&end , NULL);
        printf("%ld\t2\t\t2.1\t%.0lf\n",sno,time_diff(initial, end));
        sno++;
        gettimeofday(&end , NULL);
        printf("%ld\t2\t\t2.2\t%.0lf\n",sno,time_diff(initial, end));
           fflush(stdout); 
pthread_mutex_unlock(&lock);
        sleep(1);
    }
}

Answer 2

当检测到输出未定向到终端时，在后台将缓冲设置为块缓冲。如果您等待的时间足够长（可能生成 4096 字节的输出），那么整个输出将成批出现。

要解决这个问题，您可以在每个 printf(); 之后使用 fflush(stdout); 或强制缓冲模式在开头 行缓冲 12=].

查看 setlinebuf() 的联机帮助页了解更多信息。

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