Question 4 of 9: File I/O — Average Word Length

Question 4: File I/O — Average Word Length (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A text file whose "word" boundaries are exactly four delimiter kinds: blank, comma, period, and line start/end (the newline character serves as both end-of-line and start-of-next-line).

Find. The average number of letters per word, over the whole file.

Approach. Read the file one character at a time (no need to buffer whole lines); maintain a running "current word length" that resets to zero at every delimiter, and add it to a running total whenever a non-empty word just ended; divide total letters by total words at end of file.

1. Recognise this as a state-machine scan, not a tokenizer that needs to store words. Only the running letter-count of the word in progress is needed — never the word's actual characters — because the question asks for a length statistic, not the words themselves.

   #include <stdio.h>
   
   int is_delim(int c)
   {
       return c == ' ' || c == ',' || c == '.' || c == '\n';
   }
   
   int main(void)
   {
       char filename[256];
       long total_letters = 0, total_words = 0, cur_len = 0;
       int c;
       FILE *fp;
   
       printf("Enter file name: ");
       if (scanf("%255s", filename) != 1) return 1;
   
       fp = fopen(filename, "r");
       if (fp == NULL) {
           printf("Could not open %s\n", filename);
           return 1;
       }
   
       while ((c = fgetc(fp)) != EOF) {
           if (is_delim(c)) {
               if (cur_len > 0) {           /* a word just ended */
                   total_words++;
                   total_letters += cur_len;
                   cur_len = 0;
               }
           } else {
               cur_len++;
           }
       }
       if (cur_len > 0) {                   /* trailing word, no closing delimiter */
           total_words++;
           total_letters += cur_len;
       }
       fclose(fp);
   
       if (total_words == 0)
           printf("The file contains no words.\n");
       else
           printf("Average word length: %.4f letters (%ld words).\n",
                  (double) total_letters / total_words, total_words);
       return 0;
   }

2. Trace it on a worked example file containing the single line The quick brown fox jumps over the lazy dog. The words are The(3), quick(5), brown(5), fox(3), jumps(5), over(4), the(3), lazy(4), dog(3) — nine words, and the trailing period after "dog" closes the last word rather than being swallowed into it. $$\text{total letters} = 3+5+5+3+5+4+3+4+3 = 35, \qquad \text{total words} = 9$$ $$\boxed{\text{average word length} = 35/9 = 3.8889\ \text{letters}}$$