c - 整数与 float : Counter

Question

代码:

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

int main()
{
    FILE *fp1, *fp2;
    int ch1, ch2;
    clock_t elapsed;
    char fname1[40], fname2[40];

    printf("Enter name of first file:");
    fgets(fname1, 40, stdin);
    while ( fname1[strlen(fname1) - 1] == '\n')
    {
        fname1[strlen(fname1) -1] = '\0';
    }

    printf("Enter name of second file:");
    fgets(fname2, 40, stdin);
    while ( fname2[strlen(fname2) - 1] == '\n')
    {
        fname2[strlen(fname2) -1] = '\0';
    }

    fp1 = fopen(fname1, "r");
    if ( fp1 == NULL )
    {
        printf("Cannot open %s for reading\n", fname1 );
        exit(1);
    }

    fp2 = fopen( fname2,  "r");
    if (fp2 == NULL)
    {
        printf("Cannot open %s for reading\n", fname2);
        exit(1);
    }

    elapsed = clock(); // get starting time

    ch1  =  getc(fp1); // read a value from each file
    ch2  =  getc(fp2);

    float counter = 0.0;
    float total = 0.0;

    while(1) // keep reading while values are equal or not equal; only end if it reaches the end of one of the files
    {
        ch1 = getc(fp1);
        ch2 = getc(fp2);

    //printf("%d, %d\n", ch1, ch2);// for debugging purposes

    if((ch1 ^ ch2) == 0)
    {
       counter++;
    }

    total++;

        if ( ( ch1 == EOF) || ( ch2 == EOF)) // if either file reaches the end, then its over!
        {
            break; // if either value is EOF
        }
    }

    fclose (fp1); // close files
    fclose (fp2);

    float percent = (counter / (total)) * 100.0;

    printf("Counter: %.2f Total: %.2f\n", counter, (total));
    printf("Percentage: %.2f%\n", percent);

    elapsed = clock() - elapsed; // elapsed time
    printf("That took %.4f seconds.\n", (float)elapsed/CLOCKS_PER_SEC);
    return 0;
}

尝试比较两个大约 1.4 GB 的 .nc 文件，这些是我的结果:

$ gcc check2.c -w
$ ./a.out
Enter name of first file:air.197901.nc
Enter name of second file:air.197902.nc
Counter: 16777216.00 Total: 16777216.00
Percentage: 100.00%
That took 15.6500 seconds.

它们不可能 100% 相同，哈哈，关于为什么它似乎停在第 16777216 个字节有什么想法吗？

计数器应该是 1,256,756,880 字节

1.3 GB(1,256,756,880 字节)

我在这里下载了这个气候数据集:

ftp://ftp.cdc.noaa.gov/Datasets/NARR/pressure/

提前感谢您的帮助

Answer 1

float 数据类型仅精确到 6 位有效数字，不适用于 counter 和 total。任何浮点类型在任何情况下都是不合适的。这有很多问题，尤其是 ++ 是一个整数运算符，从 float 到 int 的隐式转换，递增，然后返回到 float 将大于 6 位的整数值失败。

我假设您选择这种类型是因为它的范围可能比unsigned int大？我建议您对这些变量使用 unsigned long long。

unsigned long long counter = 0;
unsigned long long total = 0;

...

float percent = (float)counter / (float)total * 100.0f ;