Question

引用http://www.linuxatemyram.com/，其中描述了磁盘i / o如何在内存中缓冲。我想说明这一点，并实际撤消它，以衡量将不同数量的数据写入文件所需的时间。我要做的是了解基于两种情况的i / o的运行时间：

通过mount -t tmpfs -o size=500g tmpfs /ramdisk
将数据写入/scratch/，其中900GB seagate sas硬盘格式为EXT3或XFS

我的程序在将文件写入硬盘时非常快，我确定这是因为linux操作系统正在缓冲RAM中的数据使磁盘i / o透明...因为我的程序完成后如果在提示我做rm temp_speed*这需要一分钟或更长时间才能实现。

例如，如果程序正在写入10gb，则完成写入ramdisk需要7秒，完成写入/ scratch需要20秒。但是在写入ramdisk后我可以做rm temp*并在1-3秒内完成，而如果它是在临时删除命令需要90秒才能完成。

任何人都知道如何在程序中编写下面的程序来了解磁盘i / o何时完全完成？感谢。

预先警告，如果您尝试运行此代码，则可能会很快填满您的硬盘和/或内存并导致系统崩溃。

# include <stdio.h>
# include <stdlib.h>
# include <time.h>

# define MAX_F  256

/* WARNING: running this code you risk quickly filling up */
/*          your hard drive and/or memory and crashing your system.  */

/* compile with -O0  no optimization */

long int get_time ( void )
{
   long int t;

   t = (long int) time( NULL );

   return( t );
}

void diff_time ( long int *start, long int *finish )
{
   time_t s, e;
   long int diff;
   long int h = 0, m = 0;

   s = (time_t) *start;
   e = (time_t) *finish;

   diff = (long int) difftime( e, s );

   h = diff / 3600;

   diff -= ( h * 3600 );

   m = diff / 60;

   diff -= ( m * 60 );

   printf(" problem runtime (h:m:s) = %02ld:%02ld:%02ld\n", h, m, diff );
   printf("\n");
}

int main ( int argc, char *argv[] )
{
    FILE *fp[MAX_F];
    char fname[MAX_F][64];
    long int a, i, amount, num_times, num_files;
    long int maxfilesize;
    long int start_time, end_time;
    double ff[512];     /* 512 doubles = 4096 = 4k worth of data */


    if ( argc != 3 )
    {
       printf("\n   usage: readwrite_speed  <total amount in gb> <max file size in gb>\n\n");
       exit( 0 );
    }

    system( "date" );

    amount = atol( argv[1] );
    maxfilesize = atol( argv[2] );

    if ( maxfilesize > amount )
       maxfilesize = amount;

    num_files = amount / maxfilesize;

    if ( num_files > MAX_F )
    {
       printf("\n   increase max # files abouve %d\n\n", MAX_F );
       exit( 0 );
    }

    num_times = ( amount * 1024 * 1024 * 1024 ) / 4096;

    num_times /= num_files;

    printf("\n");
    printf("   amount = %ldgb, num_times = %ld, num_files = %ld\n", amount, num_times, num_files );

    printf("\n");

    for ( i = 0; i < num_files; i++ )
       sprintf( fname[i], "temp_speed%03d", i );

    start_time = get_time();

    for ( i = 0; i < num_files; i++ )
    {
       fp[i] = fopen( fname[i], "wb" );
       if ( fp[i] == NULL )
           printf("   can't write binary %s\n", fname[i] );
    }

    for ( i = 0; i < 512; i++ )
       ff[i] = rand() / RAND_MAX;

    /* 1 gb = 262,144 times writing 4096 bytes */

    for ( a = 0; a < num_times; a++ )
    {
       for ( i = 0; i < num_files; i++ )
       {
          fwrite( ff, sizeof( double ), 512, fp[i] );
          fflush( fp[i] );
       }
    }

    for ( i = 0; i < num_files; i++ )
       fclose( fp[i] );

    end_time = get_time();

    diff_time( &start_time, &end_time );

    system( "date" );
}

Answer 1

首先，看看here 您无法通过100％保证知道磁盘上的数据您可以使用fsync告诉内核将数据刷新到磁盘。但是磁盘和磁盘控制器都有自己的缓存，所以即使内核也不能总是知道它什么时候完成。

C代码测量linux中的磁盘写入时间并处理RAM缓冲

1 个答案: