无法读取数字字符频率

Question

#include <stdio.h>

int main() {
    FILE *fb;
    char data[255];
    int c=0;
    int count[75] = {0};
    fb = fopen("Input.txt", "r");
    fgets(data, 255, fb);

    /* Start finding frequency*/
    while (data[c] != '\0')
    {
        if( data[c] >= '0' && data[c] <= 'z')
        count[data[c] - 'a']++;

        c++;
    }    
    for (c = 0; c < 75; c++)
    {
        /** Printing only those characters
            whose count is at least 1 */

        if (count[c] != 0)
            printf("%c occurs %d times in the entered string.\n",c+'a',count[c]);
    }

    return 0;
}

示例输入文件:“Fred Fish 12345678”

我能够处理输入文件中的空格，但程序无法读取大写字母和数字字符的频率。我可以在程序中更改哪些内容来帮助解决问题。阅读频率后，我的计划是保存文件，以便我可以使用 Huffman 进行压缩

Answer 1

#include <stdio.h>

int main() {
    FILE *fb;
    char data[255];
    int c = 0;
    int count[75] = { 0 };
    fb = fopen("Input.txt", "r");
    fgets(data, 255, fb);

    /* Start finding frequency*/
    while (data[c] != '\0')
    {
        if (data[c] >= 'a' && data[c] <= 'z') // here you check normal letters
            count[data[c] - 'a']++;
        else if (data[c] >= 'A' && data[c] <= 'Z')
            count[data[c] - 'A' + 26]++; // Capital letters will be stored after lower cases
        else if (data[c] >= '0' && data[c] <= '9')
            count[data[c] - '0' + 51]++; // Numbers will be stored after capital letters
        c++;
    }

    // count[] is initialized as following : 
    // From count[0] to count[25] == Occurence of low case characters
    // From count[26] to count[51] == Occurence of capital characters
    // From count[52] to count[61] == Occurence of numbers from 0 to 9

    for (c = 0; c < 61; c++)
    {
        /** Printing only those characters
        whose count is at least 1 */
        if (count[c] != 0) {
            if (c < 26)
                printf("%c occurs %d times in the entered string.\n", c + 'a', count[c]); 

                // Starting from 'a', iterating until 'z'
            else if (c < 52)
                printf("%c occurs %d times in the entered string.\n", c + 'A' - 26, count[c]);

                // Same as for low case characters
                // Substracting 26 because we already have iterated through the first 26 low case characters
                // Starting from 'A' until 'Z'

            else if (c >= 52)
                printf("%c occurs %d times in the entered string.\n", c + '0' - 51, count[c]);

                // Same as for characters
                // Substracting 51 because we already have iterated through low cases and capital characters
                // Starting from '0' until '9'
        }
    }

    return 0;
}

问题是，考虑到 ascii table ，您通过减去等于的 'a' 将大写字母和数字存储在数组 count 的负索引中到 ASCII 格式的 97。这应该可行，但我还无法测试它，所以要小心。

对于打印，我们对小写字符执行相同的操作，然后是大写字符，然后是数字:我们从第一个字符开始:'a'，然后 'A'，然后 '0'，使用 c 迭代直到最后一个并将其全部打印。我们使用的事实是，在 C 语言中，'a' + 1 = 'b'、'A' + 1 = 'B' 等。

对于输入 Fred Fish 12345678，此代码的输出为:

d occurs 1 times in the entered string.
e occurs 1 times in the entered string.
h occurs 1 times in the entered string.
i occurs 1 times in the entered string.
r occurs 1 times in the entered string.
s occurs 1 times in the entered string.
F occurs 2 times in the entered string.
1 occurs 1 times in the entered string.
2 occurs 1 times in the entered string.
3 occurs 1 times in the entered string.
4 occurs 1 times in the entered string.
5 occurs 1 times in the entered string.
6 occurs 1 times in the entered string.
7 occurs 1 times in the entered string.
8 occurs 1 times in the entered string.