Thread: Count word in a string.

So walk through the text - but use the variable inWord, and your count will be correct in ALL cases. You increment a word ONLY when inWord is 1, and you reach a non alphanumeric char.

You can't just count spaces - that will fail, and you can't make it NOT fail, without a variable like inWord.

Alternatively you can add your own list of characters that all can separate a word (by your definition). For example:

Code:

const char *punctuation = ".,!? \t\n";  // not the best name but..

// then you can test if your character is in the set with
if( strchr(punctuation, string[i]) )

Hey Adak! Thanks for your explanation. I got my code working.

Code:

          if((str[i]== ' ' && str[i+1] != ' ' && inWord++ )) 
  and also
         printf("\n\nThe total number of words are %d ", count); // removing the +1

Now my input: strcpy(str," aaa 111 === ");
Output: 3 words 4 spaces

wee~

Originally Posted by Subsonics

Alternatively you can add your own list of characters that all can separate a word (by your definition). For example:

Code:

const char *punctuation = ".,!? \t\n";  // not the best name but..

// then you can test if your character is in the set with
if( strchr(punctuation, string[i]) )

Thanks for that too. good idea

Originally Posted by Adak

I wouldn't recommend using strstr(). It's a fine function, but not for this - imo.

strchr(), but it's convenient if you would want to define "word separators" or what is to be considered a word your self.

True, but the OP is already using ctype.h so that's already done. Just needs the inWord flag to handle multiple spaces and other devious string concoctions.

Well I agree that isalpha() is probably a good assumption in most cases, but a word like "real-time" or "O'leary" and the like would be counted as 2 words for example. Then the text may be a C source file or something similar with other non alphabetical chars that you may want to have included. Anyway, I just left it there as an alternative.

There are problem chars (like hyphen and apostrophe) for sure, but adjusting the input address to work with strchr() would be beyond the OP wouldn't it?

After looking at the code again. I think i got a problem with it again.

Code:

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

int main()
{ 
           char str[50];
           int i=0,count=0;
           int space = 0;
           int len = 0;
           int inWord = 0;
           
           printf("Enter a string : ");
          // gets(str);
           strcpy(str,"Testing one two three four");
           printf("%s",str);
                 
           while(str[i]!='\0')
           {
           if((str[i]== ' ' && str[i+1]!= ' ' && inWord++ )) 
           count++;
           i++;        
           }    
           printf("\n\nThe total number of words are %d ", count);
           
           while(str[len]!='\0') len++;
           for(i=0;i<=len;i++)
{
           while( isspace( str[i] )) {  space++; i++; }
           {continue;}
        
}
          printf("\n\nThe total number of spaces are %d ",space);
           printf("\n\n");
           system("pause");
           return 0;
}

input: [space]Testing one two three[space]
output: 4 words
But
input:Testing one two three
ouput: 2 words

I think i'm getting confuse with the coding.

i have do a research on inword = 1; else inword = 0; that all the number and char = 1 and space = 0; how can i do it on my code?

OK thanks . i manage to solve the problem already