Count word in a string.

This is a discussion on Count word in a string. within the C Programming forums, part of the General Programming Boards category; So walk through the text - but use the variable inWord, and your count will be correct in ALL cases. ...

  1. #16
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    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.

  2. #17
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    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]) )

  3. #18
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    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~

  4. #19
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    Quote Originally Posted by Subsonics View Post
    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

  5. #20
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    Quote Originally Posted by Alexie View Post
    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~
    But you're not all set - you need to use isalpha() with each char.

    And inWord doesn't get ++, it gets set either to 0 or 1. Just incrementing inWord will keep inWord testing as TRUE (non-zero), all the time. You don't want that.

    I wouldn't recommend using strstr() (oops! strchr()). It's a fine function, but not for this - imo.
    Last edited by Adak; 01-20-2013 at 07:34 AM.

  6. #21
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    Quote Originally Posted by Adak View Post
    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.

  7. #22
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    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.

  8. #23
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    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.

  9. #24
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    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?

  10. #25
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    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.

  11. #26
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    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?

  12. #27
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    I think the problem is you are not considering the value of `inWord' in your procedure. Rather than starting to write code, consider a pseudocode algorithm and make sure it makes sense first:

    Code:
    wordCount := 0
    inWord := false
    c := (Get Next Character from Input)
    If isalpha(c):
         inWord := true
    End If
    If !isalpha(c):
        If inWord:
            wordCount := wordCount + 1
        End If
        inWord := false
    End If
    You have to continually repeat starting from line 3 until you get to the end of the input. Also for the above to work, the last character of the input must be a non-alpha character ('\0' and '\n' are non-alpha characters)

  13. #28
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    OK thanks . i manage to solve the problem already

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