C - Counting words in a string

Question

i've been trying to do a function that counts the number of words in a string in C. However, in some casas (as the one in the example) it should return 0 and not 1... any ideas of what could be wrong?

#import <stdio.h>  

int contaPal(char s[]) {          
    int r;     
    int i;     
    r = 0;      

    for (i = 0; s[i] != '\0'; i++) {          
        if (s[i] == '\n')           
            r = r + 0;               

        if (s[i] != ' ' && s[i + 1] == ' ' && s[i + 1] != '\0')             
            r++;        

        if (s[i] != ' ' && s[i + 1] == '\0') {              
            r++;        
        }       
    }          
    return r; 
}  

int main () {   
    char s[15] = { ' ', '\n', '\0' };

    printf("Words: %d \n", (contaPal(s)));
    return 0; 
}

@João Pimentel If the new line is encountered then r is increased due to this condition if (s[i] != ' ' && s[i + 1] == '\0') { . So the function will return at least 1 for the string you defined. — Vlad from Moscow
– Vlad from Moscow, Commented Mar 23, 2017 at 22:10
at if (s[i] != ' ' && s[i + 1] == '\0') { : ('\n' != ' ' && '\0' == '\0') become true. — BLUEPIXY
– BLUEPIXY, Commented Mar 23, 2017 at 22:15
Don't check blank characters one by one, you have many of them and this complicates your task(' ', '\t', '\n', '\r'), use isspace from ctype.h. — A.S.H
– A.S.H, Commented Mar 23, 2017 at 22:17

chqrlie · Accepted Answer · 2017-03-23 22:18:28Z

1

You should not treat '\n' differently from any other whitespace character.

Here is a simpler version:

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

int contaPal(const char *s) {          
    int count = 0, hassep = 1;

    while (*s) {
        if (isspace((unsigned char)*s) {
            hassep = 1;
        } else {
            count += hassep;
            hassep = 0;
        }
        s++;
    }
    return count;
}

int main(void) {   
    char s[] = " \n";

    printf("Words: %d\n", contaPal(s));
    return 0; 
}

answered Mar 23, 2017 at 22:18

chqrlie

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Comments

Vlad from Moscow · Accepted Answer · 2017-03-23 22:23:02Z

I suppose that the word is any sequence of characters excluding white space characters.

Your function returns 1 because for the supplied string when the new line character is encountered the variable r is increased due to this condition

    if (s[i] != ' ' && s[i + 1] == '\0') {              
        r++;        
    }

So the function implementation is wrong.

It can be defined the following way as it is shown in the demonstrative program

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

size_t contaPal( const char s[] ) 
{
    size_t n = 0;

    while ( *s )
    {
        while ( isspace( ( unsigned char )*s ) ) ++s;
        n += *s != '\0';
        while ( *s && !isspace( ( unsigned char )*s ) ) ++s;
    }

    return n;
}  

int main(void) 
{
    char s[] = { ' ', '\n', '\0' };

    printf( "Words: %zu\n", contaPal( s ) );

    return 0;
}

Its output as you expect is

Words: 0

ryyker · Accepted Answer · 2017-03-23 22:52:24Z

1

A simple illustration using existing character test functions:

int main(void)
{
    int cnt = 0;
    int numWords = 0;
    BOOL trap = 0; //start count only after seeing a word
    char *sentence = "This is a sentence, too long.";       
    //char *sentence2 = "      ";//tested for empty string also


    while (*sentence != '\0') 
    {
        if ( isalnum (*sentence) ) //word is found, set trap and start count
        {
            sentence++;  //alpha numeric character, keep going
            trap = 1;
        }
        else if ( (( ispunct (*sentence) ) || ( isspace(*sentence) )) && trap)
        {  //count is started only after first non delimiter character is found
            numWords++;
            sentence++;
            while(( ispunct (*sentence) ) || ( isspace(*sentence) ))
            { //handle sequences of word delimiters
                sentence++;
            }
        }
        else //make sure pointer is increased either way
        {
            sentence++;
        }

    }
    return 0;
}

edited Mar 23, 2017 at 22:52

answered Mar 23, 2017 at 22:34

ryyker

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6 Comments

ryyker Over a year ago

@VladfromMoscow - I did not implement a function, it is a code snippet illustrating how to use additional char test functions ispunct and isspace. To which funcntion are you referring please. I did test this in a small main function by the way.

Vlad from Moscow Over a year ago

In any case the code snippet is a part of a function is not it? It is wrong.

A.S.H Over a year ago

What would be the outcome for a sentence containing only whitespaces?

ryyker Over a year ago

@VladfromMoscow - still not sure to what you are referring, but re-wrote to include main(), surrounding rest of snippet. Does this address your concern?

ryyker Over a year ago

@A.S.H - ah! thank you. Mis-count by one. I need to test for at least one non delimiter before starting the count.

|

Alex · Accepted Answer · 2017-03-23 22:15:11Z

0

The line:

    if (s[i] != ' ' && s[i + 1] == ' ' && s[i + 1] != '\0')             
        r++;

Exactly matches the case when you look on '\n'.

You should use if ... else if ....

answered Mar 23, 2017 at 22:15

Alex

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