重定向運算符不在unix中工作？

Question

我正在運行一個程序，它在控制檯上打印所有的printf語句。 但是當我嘗試使用'>'將它們重定向到任何文件時，文件被創建，但文件中沒有輸出程序。 請幫助 當我運行在控制檯下面的代碼：

#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); 
    } 
} 

它給我下面的輸出：

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 

但是當我做./a.out> b.txt 我不在控制檯上以及在文件中沒有任何輸出

Answer 1

，緩衝設置爲塊緩衝幕後。 如果您等待了足夠長的時間（可能需要生成4096個字節的輸出），那麼整個輸出將大量出現。

爲了解決這個問題，您可以使用每printf();後fflush(stdout);或在年初setlinebuf(stdout);迫使緩衝模式明確地行緩衝。

查看setlinebuf()的手冊頁以獲取更多信息。

Answer 2

我將給出一些指示。閱讀Linux Programming Book，瞭解真正發生的事情。

我在您的代碼的相關printf()陳述後添加了fflush(stdout)，即line 32,65,85。

完成此操作後，您將獲得所需的輸出。現在試着瞭解爲什麼會出現這種行爲，爲什麼需要顯式fflush()？我認爲應該從問題的海報努力來理解這個問題，併發布Minimal Complete Verifiable Example

全碼：當檢測到輸出沒有定向到終端

#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); 
    } 
}