Simplest way to print this array

Question

What this code does simply is to break up a sentence into individual word, for example: you input My name is John, it returns:

My
name
is
John

I'll like to know if there's any better way to rewrite this?

int main() {
    int w_size = 0;
    bool check_bool = false;
    char l_str[81];
    char *ptr_to_word[81];

    for (char *res_p = &(l_str[0]); *res_p != '\0'; res_p++) {
        if ((*res_p != '.') && (*res_p != ',') && (*res_p != ' ') && (check_bool == false)) {
            ptr_to_word[w_size] = res_p;
            w_size++;
            check_bool = true;
        }
        if (((*res_p == '.') || (*res_p == ',') || (*res_p == ' ')) && (check_bool == true)) {
            check_bool = false;
        }
    }

    if (w_size == 0) {
        printf("no solution");
    } else {
        for (int i = 0; i < w_size; i++) {
            char *a = ptr_to_word[i];
            while ((*a != ',') && (*a != '.') && (*a != '\0') && (*a != ' ')) {
                printf("%c", *a);
                a++;
            }
            printf("\n");
        }
    }
    return 0;
}

Look up the `strtok` function, it will help you achieve this goal. — rtx13, Apr 16 '20 at 03:32
Does this answer your question? [Split string with delimiters in C](https://stackoverflow.com/questions/9210528/split-string-with-delimiters-in-c) — Marco Bonelli, Apr 16 '20 at 03:44
how is the posted code to attain the sentence that is to be split into words? — user3629249, Apr 16 '20 at 14:48
@Enterprevic: you can accept one of the answers by clicking on the grey checkmark below its score. — chqrlie, May 03 '20 at 22:53

user3629249 · Answer 1 · 2020-04-17T19:51:58.683

the following proposed code:

prompts the user for the sentence to be divided into words
cleanly compiles
performs the desired functionality

And now, the proposed code: (EDIT per chqrlie)

#include <stdio.h>
#include <string.h>

#define MAX_BUF_LEN 1024
#define MAX_WORDS 100


int main( void )
{
    char buffer[ MAX_BUF_LEN ] = {0};
    char *words[ MAX_WORDS ] = {NULL};

    printf( "%s\n", "Please enter a sentence to be divided into words" );

    if( fgets( buffer, sizeof( buffer ), stdin ) )
    {
        size_t wordCount = 0;
        char *token;

        token = strtok( buffer, ",. " );
        while( wordCount < MAX_WORDS && token )
        {
            words[ wordCount ] = token;
            wordCount++;
            token = strtok( NULL, ",. " );
        }

        for( size_t i = 0; i < wordCount; i++ )
        {
            printf( "%zu: %s\n\n", i+1, words[i] );
        }
    }
}

Here is the results of a typical run of the proposed code:

Please enter a sentence to be divided into words
This is a sentence to be divided into words
1: This

2: is

3: a

4: sentence

5: to

6: be

7: divided

8: into

9: words

regarding: `for( size_t i = 0; words[i]; i++ )` the array of `words[]` is initialized to all NULL, so the second parameter in the `for()` statement will cause the loop to exit when it encounters an entry in `words[]` that contains NULL — user3629249, Apr 16 '20 at 21:58
Yes this works, but it excludes capital letters, any fix around that? and also it doesn't take input of punctuation marks as its skips them, is it possible to take all as input then just skips out when printing output? Thanks for the solution — Enterprevic, Apr 17 '20 at 05:16
@chqrlieforyellowblockquotes, I wish you had put it that way to begin with. I'll modify my answer. — user3629249, Apr 17 '20 at 19:50

chqrlie · Answer 2 · 2020-04-17T21:21:58.427

If you dont need to store the words into an array, you can output them directly:

#include <stdio.h>
#include <string.h>

int main() {
    char str[81];

    printf("Enter string: ");
    if (fgets(str, sizeof str, stdin)) {
        int pos = 0, len, index = 1;
        for (;;) {
            /* skip initial separators */
            pos += strspn(str + pos, ",.\n ");
            if (str[pos] == '\0')
                break;
            /* compute the length of the word */
            len = strcspn(str + pos, ",.\n ");
            printf("%d: %.*s\n", index++, len, str + pos);
            pos += len;
        }
    }
    return 0;
}

Simplest way to print this array

2 Answers2