number of words in a string

hi everyone, i made this code but i'm not sure if it's 100% correct

Code:

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

  int count(char x[]){

          int sum=0,j,cw=0;

          for (j=cw;j<=strlen(x)-1;j++){

          while(x[j]!=' '){
          cw++;
          j++;
            }

        if (x[j+1]!=' ' && !isdigit(x[j+1])){
                          sum++;
                  }
        }

        if (x[0] == ' ' || isdigit(x[0])){
                        sum--;
              }

  return sum;

}

int main(void){

    char x[1000] = " a b c d ef g 90123012031203 hi jjj 0210312093091230123";

    printf("%d\n",count(x)); //the result is 8

    getchar();

  return 0;

}

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it prints me the correct results for different input, but is there any other better/smarter way to do this?

thanks

In most cases, your correct answer would be 10 with your input. Not that 'c' is a word, but unless you're checking with a dictionary, you have to count it as one.

The indentation is very poor, but ++j, being done separately inside a for loop with j as the counter, is usually an error, right there. Why would you do that?

So "90210" isn't going to count as a word? "Fahrenheit 451" is only going to be one word? Better re-think that. Adjacent numbers have to be counted as a word if they have a space before them, and either punctuation (including newlines), or a space, after them.

You've Rube Goldberg'd this. All you need is a single for or while loop, with a few if statements inside.

That's it.

Quote:

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Originally Posted by Adak

In most cases, your correct answer would be 10 with your input. Not that 'c' is a word, but unless you're checking with a dictionary, you have to count it as one.

The indentation is very poor, but ++j, being done separately inside a for loop with j as the counter, is usually an error, right there. Why would you do that?

So "90210" isn't going to count as a word? "Fahrenheit 451" is only going to be one word? Better re-think that. Adjacent numbers have to be counted as a word if they have a space before them, and either punctuation (including newlines), or a space, after them.

You've Rube Goldberg'd this. All you need is a single for or while loop, with a few if statements inside.

That's it.

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thanks for the reply, it was my fault not to say that we dont count numbers as words

new line characters, commas etc and punctuations will not be included in the string

also, why do you say that j++ is an error? i use it in order to make less checkings in the string

can you tell me exactly what you mean? maybe some extra code would help me understand

also for different inputs i get

" 1234 Hello world 2340923042 how are you " = 5


" 1234 Hello world asdasdasdasd asdas das da sd asd 2340923042 how are you " = 11

maybe you should take in consideration the function strtok

Quote:

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Originally Posted by jackhasf

thanks for the reply, it was my fault not to say that we dont count numbers as words

new line characters, commas etc and punctuations will not be included in the string

also, why do you say that j++ is an error? i use it in order to make less checkings in the string

can you tell me exactly what you mean? maybe some extra code would help me understand

---

I can see that you don't count numbers as words. My point is that the book title "Fahrenheit 451" has more than one word in it's title.

Obviously, newlines, etc., are not included in the string, but they mark the end of a word.

"...word." <<== for example

j++ is the counter in the for loop. Your logic is nearly always goofed if you are messing with the counter, inside a for loop. And since your word count is wrong... :)

You have two states for your char counter, it's either inside a word, or it's outside of a word, at any instant.

When the array[i] = any char and your state is currently outside, then you know you need to switch the state to inside, and add a word to the word count.

When your array[i] = a space, a punctuation, or a newline, then you need to switch the state from inside, to outside again.

These tests can be done with just a few if statements, right inside your ONE for loop.

STOP **THINKING**, Dammit! Just listen and **DO** it. The answer is so clear, it will hit you right in the chops!

This is a simple simon exercise. You don't need strtok() or any other library. IT"S TOO SIMPLE, for gawd's sake!

Remember: one for loop, and a few if statements. That's ALL!!

I'm waiting for the day when a guy posts his pet chicken pecking out this program, on YouTube. :p

thanks again for the reply

Quote:

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Originally Posted by Adak

I can see that you don't count numbers as words. My point is that the book title "Fahrenheit 451" has more than one word in it's title.

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True, but, i don't want to count numbers, that's why I did what I did and in this program if you have the fahrenheit 451 input, you will get 1 as a result, which is the correct result.

Quote:

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Originally Posted by Adak

Obviously, newlines, etc., are not included in the string, but they mark the end of a word.

---

As I know '\0' marks the end of a string, and i dont really need to know where it is cause I use strlen() function.

Quote:

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Originally Posted by Adak

"...word." <<== for example

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as i said before, string can't and won't have punctuation, it's what the problem states.

Quote:

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Originally Posted by Adak

exercice
j++ is the counter in the for loop. Your logic is nearly always goofed if you are messing with the counter, inside a for loop. And since your word count is wrong... :)

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why is it wrong? I get the correct results, I can't really understand what's wrong in my code, since for every kind of input, I get the correct answer, I had a sheet of paper with me while doing this, and I can say I'm 99% sure that my logic is not goofed.

talking about errors of your code:
in

Code:

---

          while(x[j]!=' '){
          cw++;
          j++;
            }

---

you should make a check that j will not go over the lenght of the string

there are probabilities that you would get a segmentation fault or erroneous output

same check when you write

Code:

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if (x[j+1]!=' ' && !isdigit(x[j+1])){

---

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adak already give a better solution, using a single loop, checking every step these things:
1 - if the actual character is a letter from 'a' to 'z' or from 'A' to 'Z', if not continue to loop
2 - increment the words_count every time you enter in a word (a series of letters of the alphabet)
3 - incrementing the index_loop and checking is not out of bound

One really, really easy way to think of a word is just as non-whitespace characters surrounded by spaces (or the ends of the string). With that in mind, I think something as simple as this might work for you . . .

Code:

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int word_count(const char *string) {
    int wc = 0;
    int x = 0;
    for(;;) {
        /* skip whitespace, if any */
        while(string[x] == ' ') x ++;

        /* stop if we've reached the end of the string */
        if(!string[x]) break;

        /* skip over the word */
        while(string[x] != ' ' && string[x]) x ++;

        /* skip whitespace, if any */
        while(string[x] == ' ') x ++;
       
        /* increment word count */
        wc ++;
    }
    return wc;
}

---

[Coloured with codeform.]

Of course, there are other ways (perhaps even simpler ones). But hopefully that will give you some ideas. Note that I haven't test it, so no guarantees. :)

[edit] As others have suggested, it's probably better not to mess with the loop variable inside the loop, as I did (trying to make the code a little similar to yours). Another easy way to do this is to remember the character you saw during the last iteration of the loop (specifically, whether it was a space or not). Then just take note when you change from space to non-space, and from non-space to space. In the former case, you have the beginning of a word, so count it. For example:

Code:

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char last_char = ' ';
for(x = 0; x < length; x ++) {
    if(last_char == ' ' && string[x] != ' ') {
        wc ++;
    }
    last_char = string[x];
}

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[/edit]

If you don't want to learn a simpler, slightly more efficient, and more robust way of doing this, then I don't understand why you posted here.

You've got a program you're happy with. It's giving you the right answers if you're not counting numbers. You don't want to change anything.

OK.

As the saying goes: You can't taste my tea, without first emptying your cup.

Quote:

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Originally Posted by Adak

If you don't want to learn a simpler, slightly more efficient, and more robust way of doing this, then I don't understand why you posted here.

You've got a program you're happy with. It's giving you the right answers if you're not counting numbers. You don't want to change anything.

OK.

As the saying goes: You can't taste my tea, without first emptying your cup.

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ofcourse i wanted to change things, but you kept saying that my code and thinking was wrong without telling me why

Quote:

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Originally Posted by jackhasf

ofcourse i wanted to change things, but you kept saying that my code and thinking was wrong without telling me why

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I wanted you to *DISCOVER* what I was showing you - not be told, not by analyzing a function I wrote for you. Just take a few pointed suggestions, and with what you already know, you'd run right into it.

And be a bit amazed, perhaps.

If I had said your code and algorithm was great, why would you want to change it? What would be my argument to support that change? :)

That makes no sense to me.

I believe you should have taken my word for something - and gave it a shot.

Which is all OK, by the way. You don't know me, and there's no reason you should trust me to give you good advice on your program.

Be well.