Question

我在对它们进行排序后加入两个文件，并评估我需要获得平均时间的时间。但根据我的编码，它会在几秒钟内显示，因此执行速度足以让结果在1秒内完成，因此无需获取平均值。这就是为什么我想以毫秒为单位打印时间，以便我可以采用这些时间并计算平均时间。这是我的代码 -

#include <sys/stat.h>   // S_IRUSR
#include <time.h>       // time_t time
#include "mergesort.h"

typedef struct {
int fd; //file description ID
int type; //the schema type of file
page_t *page; //page buffer which use to store records
} file_t;


/*! \fn file_t *file_init(char *file_path, int type, int page_size)
 *  \brief initialize the file structure with inputted parameters
 *  \param file_path the path of need to handled file     
 *  \param type the type of file. 
 *         1 means the file formated with characters schema
 *         2 means the file formated with guilds schema
 *  \param page_size the size of page, the page will use to buffer the 
 *         record which read from file
 *  \return not NULL the file structure
 *          NULL means cannot construct the file structure
 */
file_t *file_init(char *file_path, int type, page_t *page)
{
file_t *file = calloc(1, sizeof(file_t));

char sorted_file[1024] = {0};
snprintf(sorted_file, 1023, "%s.sorted", file_path);
file->fd = open(sorted_file, S_IRUSR);
file->type = type;
file->page = page;
return file;
}

int merge_sort(pass_t *pass, int page_size, int page_num, char *file)
{
int ret = OK;
if (pass->pages == NULL)
{
    pass->pages = calloc(pass->page_num, sizeof(page_t));
}

// parse input
pass->page_size = page_size;
pass->page_num = page_num;
pass->input_file = file;
pass->output_file = "/tmp/merge_tmp";
pass->type = 1;
if (strstr(pass->input_file, "guilds") != NULL)
{
    pass->type = 2;
}


// pass 0: split the original into sorted file pieces
pass_0(pass);

// pass n: merge the file pieces into a sorted unique file
return pass_n(pass);
}

/*! \fn int main(int argc, char *argv[])
*  \brief the entry of program
*  \param argc the number of input parameters
*  \param argv the list of input parameters
*         1st is program name
*         2nd is the optional flag to indicate ouput matched record to screen 
*         2nd or 3rd is the record size of page
*         3rd or 4th is the page size
*         4th or 5th is one file which need to be merged
*         5th or 6th is another file which need to be merged
*  \return 0 indicate the operation is success
*          other indicate the operation is failed
*/
int main(int argc, char *argv[])
{
if (argc < 4) 
{
    fprintf(stderr, 
            "Usage: bin/sortmerge [-o] <pagesize> <buffers> <path>\n");
    exit(1);
}

int param_id = 1;
int out_flag = 0;
if (strcmp(argv[param_id], "-o") == 0)
{
    out_flag = 1;
    param_id++;
}

int page_size = atoi(argv[param_id++]);
int page_num = atoi(argv[param_id++]);
int total_buf = page_size * page_num;

int ret = OK;
int tuples_num = 0;
time_t start = time(NULL);

pass_t file1_pass = {0};
merge_sort(&file1_pass, page_size, page_num, argv[param_id++]);

pass_t file2_pass = {0};
// reuse the allocated buffer
file2_pass.pages = file1_pass.pages;
merge_sort(&file2_pass, page_size, page_num, argv[param_id++]);

// open sorted file
file_t *chars_file = NULL;
file_t *guilds_file = NULL;
if (strstr(file1_pass.input_file, "guilds") != NULL)
{
    chars_file = file_init(file2_pass.input_file, 1, &file2_pass.pages[0]); 
    guilds_file = file_init(file1_pass.input_file, 2, &file1_pass.pages[1]);
}
else
{
    chars_file = file_init(file1_pass.input_file, 1, &file1_pass.pages[0]); 
    guilds_file = file_init(file2_pass.input_file, 2, &file2_pass.pages[1]);
}

record_t *guild_record = NULL;
record_t *char_record = NULL;
PAGE_BEFORE(chars_file->page, page_size);
PAGE_BEFORE(guilds_file->page, page_size);
do
{
    // load and popup record from guilds
    page_load(guilds_file->fd, guilds_file->page, guilds_file->type, 
            guilds_file->type);
    guild_record = page_pop_record(guilds_file->page);
    if (guild_record == NULL)
    {
        //no record then exit
        break;
    }

    do
    {
        // load popup record from characters
        page_load(chars_file->fd, chars_file->page, chars_file->type, 
                chars_file->type);
        char_record = page_pop_record(chars_file->page);
        if (char_record != NULL 
                && guild_record->guild_id == char_record->guild_id)
        {
            // records matched with guild_id
            tuples_num++;
            if (out_flag == 1)
            {
                // output matched record information
                printf("%d,%s,%d,%d,%d,%s\n", char_record->guild_id, 
                        char_record->info.character.cname,
                        char_record->info.character.team,
                        char_record->info.character.level,
                        char_record->info.character.cid,
                        guild_record->info.guild.gname);
            }
        }

        // if not find bigger than guild tuple record, 
        // continuously popup record from characters
    } 
    while (char_record != NULL 
            && guild_record->guild_id >= char_record->guild_id);

}
while (guild_record != NULL && char_record != NULL);

printf("Number of tuples: %d\nTime: %d seconds\n", 
        tuples_num, (time(NULL) - start));
return 0;
}

Answer 1

使用clock（）而不是time（）。您应该能够将时间（NULL）更改为clock（）。 clock_t也可以替换time_t。

如何用C表示以毫秒为单位的时间

1 个答案: