c - 为什么不同的测试有完全相同的数字？

Question

为什么缓冲 0.030000 秒与“更好”缓冲 0.030000 秒相同？如果 4 倍大的线条尺寸不会改变时间，我怎样才能加快速度？

测试

$ ./a.out 
Unbuffered: 0.770000 seconds
Buffered: 0.030000 seconds
Better buffered: 0.030000 seconds

代码

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define SIZE 1024 /* read 1024 bytes at a time */
#define betterSIZE 4096 /* read a better size at a time */

int copy() /* input2output ie anything to anything */
{
    char buf[betterSIZE];
    int n;
    while ((n = read(0, buf, betterSIZE)) > 0)
    write(1, buf, n);
    return 0;
}

int main(int argc, char **argv)
{

    /* copy(); */

    /* make the names known */

    void info(char file_name[]);
    void buffered(char file_name[]);
    void better_buffered(char file_name[]);

   /* test */

    clock_t toc;
    clock_t tic = clock();
    info("coreutils_8.13.orig.tar.gz"); 
    info("coreutils_8.13.orig.tar.gz"); 
    info("coreutils_8.13.orig.tar.gz"); 
    info("coreutils_8.13.orig.tar.gz"); 
    info("coreutils_8.13.orig.tar.gz"); 
    toc = clock();
    printf("Unbuffered: %f seconds\n", (double)(toc - tic) / CLOCKS_PER_SEC);
    tic = clock();    
    buffered("coreutils_8.13.orig.tar.gz"); 
    buffered("coreutils_8.13.orig.tar.gz"); 
    buffered("coreutils_8.13.orig.tar.gz"); 
    buffered("coreutils_8.13.orig.tar.gz");
    buffered("coreutils_8.13.orig.tar.gz"); 
    toc = clock();
    printf("Buffered: %f seconds\n", (double)(toc - tic) / CLOCKS_PER_SEC);
    tic = clock();    
    better_buffered("coreutils_8.13.orig.tar.gz"); 
    better_buffered("coreutils_8.13.orig.tar.gz"); 
    better_buffered("coreutils_8.13.orig.tar.gz"); 
    better_buffered("coreutils_8.13.orig.tar.gz");
    better_buffered("coreutils_8.13.orig.tar.gz"); 
    toc = clock();
    printf("Better buffered: %f seconds\n", (double)(toc - tic) / CLOCKS_PER_SEC);
    return 0;
}

void info(char file_name[])
{
    int ch;
    FILE *fp;
    fp = fopen(file_name,"r");
    // read mode
    if (fp == NULL)
    {
        perror(file_name);
        exit(EXIT_FAILURE);
    }
    while ((ch = fgetc(fp)) != EOF)
    {
        //putchar(ch);
    }
    fclose(fp);
}

void buffered(char file_name[])
{
    char buf[SIZE];
    FILE *fp;
    size_t nread;
    fp = fopen(file_name, "r");
    if (fp) {
        while ((nread = fread(buf, 1, sizeof buf, fp)) > 0)
    {
            //fwrite(buf, 1, nread, stdout);
    }
        if (ferror(fp)) {
            /* to do: deal with error */
        }
        fclose(fp);
    }
}


void better_buffered(char file_name[])
{
    char buf[betterSIZE];
    FILE *fp;
    size_t nread;
    fp = fopen(file_name, "r");
    if (fp) {
        while ((nread = fread(buf, 1, sizeof buf, fp)) > 0)
    {
            //fwrite(buf, 1, nread, stdout);
    }
        if (ferror(fp)) {
            /* to do: deal with error */
        }
        fclose(fp);
    }
}

Answer 1

fread 已经缓冲数据，即在内部以通常大小为 4K 或 8K 的 block 读取数据(确切的默认缓冲区大小由实现定义)。出于这个原因，将您的读取从 1K 切换到 4K 没有任何作用，因为它只会减少 memcpy 操作的数量，这些操作足够便宜，不会在统计噪声以上的基准中注册。

要影响stdio 缓冲区的大小，请查看setbuf。 .

此外，您要更改的不是行大小，因为您不是逐行读取文件，而是 block 大小。