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c - 用 C 计算文件中的字母

问题描述

我正在尝试创建一个程序，该程序从文件中读取并计算文件中每个字母字符的出现次数。以下是我到目前为止的内容，但是返回的计数（存储在 counters 数组中）比预期的要高。

  void count_letters(const char *filename, int counters[26]) {
    FILE* in_file = fopen(filename, "r");
    const char ALPHABET[] = "abcdefghijklmnopqrstuvwxyz";
    if(in_file == NULL){
        printf("Error(count_letters): Could not open file %s\n",filename);
        return;
    }
    char line[200];
    while(fgets(line, sizeof(line),in_file) != NULL){ //keep reading lines until there's nothing left to read
        for(int pos = 0; pos < sizeof(line); pos++){//iterate through each character in line...
            if(isalpha(line[pos])){//skip checking and increment position if current char is not alphabetical
                for(int i = 0; i < 26; i++){//...for each character in the alphabet
                    if(tolower(line[pos]) == tolower(ALPHABET[i]))//upper case and lower case are counted as same
                        counters[i]++;    // increment the current element in counters for each match in the line
                }
            }
        }
    }
    fclose(in_file);
    return;
}

标签： cfile

解决方案

我的两分钱用于一个更简单的解决方案（你有很多循环；））。在大多数情况下，首选逐行读取输入，但由于您只是在这里计算字符，我认为这不是其中之一，最终会增加复杂性。该答案还假定ASCII 字符编码，正如另一个答案的评论中所述，C 标准不保证这一点。您可以根据需要进行修改以char ALPHABET获得最终的便携性

#include <stdio.h>
#include <ctype.h>

#define NUM_LETTERS 26

int main(void)
{
    FILE* in_file = fopen("/path/to/my/file.txt", "r");
    if (in_file == NULL) exit(-1);

    unsigned charCounts[NUM_LETTERS] = {0};
    int curChar;
    // rather than reading line-by-line, read one character at a time
    while ((curChar = fgetc(in_file)) != EOF)
    {
        // only proceed if it is a letter
        if (isalpha(curChar))
        {
            // this is bad if not using ASCII, instead you'd need another
            // loop to check your ALPHABET, but increment the count here
            (charCounts[tolower(curChar) - 'a'])++;
        }
    }

    // print out the results
    for (int i=0; i<NUM_LETTERS; i++)
    {
        // 'A'+i also assumes ASCII encoding
        printf("%c: %u\n", 'A'+i, charCounts[i]);
    }
}

演示使用stdin而不是文件。

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