程序查找字长统计

Question

I am expected to make a program to calculate and display statistics about the length of words in a text file. I have been provided the following file

int readFile(const char fName[], char textStr[]){
    FILE *fPtr;
    char ch;
    int size = 0;

    if ((fPtr = fopen(fName, "r")) == NULL) {
        fprintf(stderr, "Error, failed to open %s: ", fName);
        perror("");
        return 1;
    }

    while ((ch = fgetc(fPtr)) != EOF) {
        if (size >= MAX_FILE - 1)
            break;
        textStr[size++] = ch;
    }

    textStr[size] = '\0';

    return size;
}

I was able to verify that I can access the file using the following code

int main() {
    char str[MAX_FILE];
    int len = readFile("test.txt", str);
    if (len == -1) {
        printf("An error occurred\n");
    } else {
        printf("file read");
    }
}

File test.txt contains

The quick brown fox jumps over the lazy dog

What I want to do is to get the contents of test.txt and find the length of each word in it something like:-

1 letter words- 0
2 letter words - 0
3 letter words - 3
4 letter words -4

and so on...

Answer 1

As a fellow new contributor, I'm going to give you a break and try to answer the question you didn't ask. ;)

I believe the question is "how to proceed". This is going to be a long answer as I will try be very detailed since you seem to be a newbie. Hopefully this will help you or maybe someone else.

The trick is to take a word problem and convert it into a mathematical solution. The best way to do this is to write "pseudocode". (See Wikipedia for more information, if you need to.) I'm going to give you some pseudocode at the end, but since this appears to be a homework assignment, please try to write your own pseudocode first. If you read the pseudocode and it still doesn't help, I can post my solution later. (I'm not a great programmer so it might not be the best program. And it took way overlong to come up with it.)

First things first: There appears to be a typo in the code you posted. In the source code you were provided, the problem is the return 1 statement if the file isn't found. That should be return -1 , because what would happen if you had a test file that had exactly 1 letter? The code wouldn't work correctly.

Now, to first convert the word problem you were given: You need to have an array of word counts to keep track of 1-letter, 2-letter, etc. words. Now according to this the longest word in the English dictionary is 45 letters. So, in theory, you would need to have an array of 45 elements of wordCounts . You can shorten this as required.

Now to process your str variable, you need a while statement to go through one character at a time. Since the characters in the string go from element 0 through one less that the len variable, you need to code the while accordingly.

Within that while , you need another while . This while needs to count up the wordLength one character at a time, until you see a blank or the trailing '/0' character of str . To do this, you initialize the wordLength to zero right before the second while. Then add 1 to the wordLength for each character you count and increment your subscript .

At the end of this inner while you need to accumulate your wordCounts. Keep in mind that your 1-letter words are going to be accumulated into element 0 of your array. So you need to adjust the wordLength - 1 array element. After that you need to increment your subscript you are using to go through your str , one character at a time.

At the end, you need to print out the wordCounts array values. Since most of the word lengths will have a value of zero, I wouldn't print these. Unless you set the maximum length of the wordCounts array to something like 10, instead of 45. You want a for loop to go through your wordCounts array, and do something like this: printf("%2d letter words = %d", ..., ...); . Keep in mind your 1-letter words are going to be in element 0;

That is a very detailed version of a word problem that is the solution to the problem of "count the number of words that the phrase has from 1-letter words to x -letter words".

Here is the pseudocode I came up with, after coding my solution. It is a little more detailed than normal pseudocode would be. (Personally, I abbreviate all variable names and use Pascal case, but that's just me.)

Declare a numeric array of wordCounts and a subscript .

For each element of wordCounts, zero out the number of words or the code won't work right.

Reinitialize subscript to zero.

As long as ( while ) the subscript is less than the len , continue.

Initialize the wordLength to zero.

As long as the str[subscript] is not a blank or a null character, add 1 to the wordLength.

Increment the subscript.

After both while statements are complete print out the array of wordLengths, as described above.

Your done!

Now I could post the actual code that could be used to come up with this pseudocode, but it would be better if you came up with it yourself. If you try but have a bug in your code, post a new question, and I'll try to check back to answer it. Hope this helps you or someone else.