Clock() 没有按预期工作；避免IO

Question

我正在编写一个程序，一次读取 1MB 的大文件 (44GB - 63GB)，然后我对这 1MB 进行哈希处理。但是，我想看看执行这些哈希需要多长时间

我对一次读入一个 1MB 的文件需要多长时间不感兴趣，只关心哈希性能时间。目前我正在使用一个非常基本/通用的哈希函数

关于时钟开始和结束时间的任何想法？

这是我目前所拥有的:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#define HASH_PRIME 65551// prime number for hash table

// generic hash function
static unsigned short hash_Function(char *hash_1MB)
{
    unsigned short hash;
    int i = 0;
    while(hash_1MB[i]!='\0')//each char of the file name
    {
        hash += (unsigned short)hash_1MB[i];//add it to hash
        i++;
    }
    return hash%HASH_PRIME;//mod hash by table size
}

int main()
{
    struct stat fileSize;
    char *buffer;

    FILE *fp;
    clock_t start, stop;
    double duration;
    char fname[40];

    printf("Enter name of file:");
    fgets(fname, 40, stdin);
    while (fname[strlen(fname) - 1] == '\n')
    {
        fname[strlen(fname) - 1] = '\0';
    }

    // handle file, open file, and read in binary form
    fp = fopen(fname, "rb");
    if (fp == NULL)
    {
        printf("Cannot open %s for reading\n", fname);
        exit(1);
    }

    stat(fname, &fileSize);
    size_t size = fileSize.st_size;
    printf("Size of file: %zd\n", size);

    buffer = (char*) malloc(sizeof(*buffer)*1000*1000);

    unsigned long long counter = 0;
    // read in 1MB at a time // & start timing how long it takes to perform the hash
    start = clock();
    clock_t total = 0;
    while (fread(buffer, sizeof(*buffer), (1<<20), fp) == (1<<20)) 
    {
    start = clock();
    hash_Function(buffer);  
    counter++;
    total += (clock() - start);
    }

    //free(buffer);

     fclose (fp); // close files

     duration = (double)((stop - start)/CLOCKS_PER_SEC);

     printf("Counter: %llu\n", counter); // how many MB were hashed
     printf("Hashing took %.2f seconds\n", (float)duration);
     return 0;
}

我的结果也没有像预期的那样出来，我分析的第一个文件有 1,961,893,364 字节大，所以应该至少有 1,961MB 被散列

但是当我打印出我的计数器来检查正确数量的 MB 被散列时，我只得到 1871

这是我的结果:

$ gcc one_mb.c
$ ./a.out
Enter name of file:v.10.nc
Size of file: 1961893364
Counter: 1871
Hashing took 0.00 seconds

提前感谢您的帮助!

/////结果为 (1000*1000)

Enter name of file:v.13.nc
Size of file: 15695146912
Counter: 15695
Hashing took 18446744.00 seconds

//////1 << 20

结果

Enter name of file:v.13.nc
Size of file: 15695146912
Counter: 14968
Hashing took 18446744.00 seconds // why this long?!?!? It didn't take 30mins

/////用for循环替换while循环

// generic hash function
static unsigned short hash_Function(char *hash_1MB)
{
    unsigned short hash;
    int i;

    for(i = 0; i < (1 << 20); i++)
    {
        hash += (unsigned short)hash_1MB[i];//add it to hash
    }

    return hash%HASH_PRIME;//mod hash by table size
}

Answer 1

您需要在 while 循环中获取时间戳并保留它们的总和以避免对文件 IO 计时。

start = clock();
clock_t total = 0;
while (fread(buffer, 1<<20, (1<<20), fp) == (1<<20)) 
{
    start = clock();
    hash_Function(buffer);  
    counter++;
    total += (clock() - start);
}

请注意，我将 1000*1000 更改为 1<<20，因此它实际上是一个 MB 的大小。

还要确保至少为 1 MB 正确分配缓冲区。

buffer = (char*) malloc(1<<20);

以下计算结果为(字符大小)* 1000 * 1000 = 1000 * 1000，这是行不通的。

buffer = (char*) malloc(sizeof(*buffer)*1000*1000);

此外，当您执行 sizeof(*buffer) 时，这也会返回 char 的大小(1 字节)。查看更新后的恐惧。