c - 字数统计程序 - stdin

Question

下面为question ,

Write a program to read English text to end-of-data (type control-D to indicate end of data at a terminal, see below for detecting it), and print a count of word lengths, i.e. the total number of words of length 1 which occurred, the number of length 2, and so on.

Define a word to be a sequence of alphabetic characters. You should allow for word lengths up to 25 letters.

Typical output should be like this:

        length 1 : 10 occurrences
        length 2 : 19 occurrences
  length 3 : 127 occurrences
     length 4 : 0 occurrences
        length 5 : 18 occurrences
        ....

To read characters to end of data see above question.

---

这是我的工作解决方案，

#include<stdio.h>
int main(void){
  char ch;
  short wordCount[20] = {0};
  int count = 0;
  while(ch = getchar(), ch >= 0){
    if(ch == ' ' || ch == ',' || ch == ';'|| ch == ':'|| ch == '.'|| ch == '/'){
      wordCount[count]++;
      count=0;
    }else{
      count++;
    }
  }
  wordCount[count]++; // Incrementing here looks weird to me

  for(short i=1; i< sizeof(wordCount)/sizeof(short); i++){
    printf("\nlength %d : %d occurences",i, wordCount[i]);
  }
}

---

问题:

1)

从代码优雅的角度来看，我可以避免在 while 循环之外递增 (++) wordCount 吗？

2)

我能否根据单词大小使 wordCount 数组大小更加动态，而不是恒定大小 20？

注意:了解了 struct 但还没有学习动态结构，例如 Linkedlist

Answer 1

对于动态分配，您可以从 20 个 shorts 的空间开始(尽管问题陈述似乎要求您允许最多 25 个字符的单词):

short maxWord = 20;
short *wordCount = malloc(sizeof(*wordCount) * maxWord);

然后，当您增加 count 时，如果当前单词的长度超过动态数组中可以计算的长度，您可以分配更多空间:

} else {
    count++;
    if (count >= maxWord) {
        maxWord++;
        wordCount = realloc(sizeof(*wordCount) * maxWord);
    }
}

完成后不要忘记 free(wordCount)。

由于您不需要计算零长度单词，您可以考虑修改您的代码，以便 wordCount[0] 存储长度为 1 的单词数，依此类推。