C++ "counting words in a string"

Rico16135 · 04-23-2004, 02:48 PM

In the book I am reading the suggested exercise is to write a function to count the number of words in a string. Then include what I define as a word in my documentation. Well, I haven't done alot with strings so I'm feeling a bit overwhelmed.

For this exercise a word should be:
A series of alphabetical characters
Seperated by spaces or commas
(Any other suggestions? I could get real thorough, but its just an exercise.)
Also, I am not clear how to proceed. Creating the function is not hard, but how do I parse a string??

Be patient I'm relatively new to programming.

h/w · 04-23-2004, 03:02 PM

hi, maybe i can write something and show ya when i get home from work. i would suggest you go through this link quick on C++.

http://www.research.att.com/~bs/bs_faq2.html

it's got links outside from there, which are worth a read too.

zekko · 04-23-2004, 04:08 PM

Since this is a exercise, you should write the actual code, but it's not too hard.

Get a string from the user, then loop through the string until '\0'
When looping, look for spaces, commas, tabs, newlines (if you keep getting input until EOF) and if found, use a variable to track how many times (count++).

Hope this helps ...

Mega Man X · 04-23-2004, 04:25 PM

It has been a while since I used C++, so I won't post any real code since it will be wrong

. But search for the function len(). Len counts the number of characters(length) in a string. Create two variables:

int numberWords;
String yourString;

Then make a for-next loop:

for (int i = 0; i <= yourString.len(); i++) {
...
// inside this loop, check to see if the hole string has "space" or "comma"
// if it has, then increase numberWords by one
// numberWords = numberWords +1
}

// show the results:

cout << numberWords << end;

I think you got the idea. The code above is all wrong for sure, but it's for you to get an idea

zekko · 04-23-2004, 05:01 PM

Oh yeah, you'll probably want to make some variable to tell if you're inside a word or not. If you are inside a word and a space, comma etc comes up, now you're outside. When a character comes up thats not some kind of space, you're inside the word and can upper the amounts of counted words.

Heh, I hope this makes sense ...

itsme86 · 04-23-2004, 05:43 PM

I'll do an example for you in C. It should compile using a C++ compiler, but it's not C++ code per se

Code:

itsme@dreams:~/C$ cat countwords.c
#include <stdio.h>
#include <string.h>

#define IS_PUNC(c) ((c) == ' ' || (c) == '.' || (c) == ',')

int word_count(char *str)
{
  int count = 0;
  char *s;

  // Skip leading punctuation and white-space
  while(IS_PUNC(*str))
    str++;
  s = str;

  while(*str)
  {
    while(*s && !IS_PUNC(*s))
      s++;
    if(str - s)
      count++;
    while(IS_PUNC(*s))
      s++;
    str = s;
  }

  return count;
}

int main(int argc, char **argv)
{
  if(argc != 2)
  {
    puts("Usage: countwords <string>");
    return 1;
  }

  printf("%d\n", word_count(argv[1]));
  return 0;
}

itsme@dreams:~/C$ ./countwords "This is a test, this is only a test."
9

Rico16135 · 04-26-2004, 04:49 PM

/* Thanks all for the advice and help. I have made a version myself, but I am getting a compiler error. It would be great if ya'll could look at it. I think I've just been missing a small fundamental issue, hence my error. */

#include <string.h>
#include <iostream.h>

main() {

char user_string[100];
int words_in_string = 0;
cout << "This program takes a sentence input by the user and returns" << '\n';;
cout << " the amount of words." << '\n';
cout << "Please input now: ";
cin.getline(user_string, sizeof(user_string)); // gets the string to use

int word_count(char input_string); // calling the function
words_in_string = word_count(user_string[100]); // thought this would work

if(words_in_string == 1) {
cout << "There is " << words_in_string << " word in the string." << '\n';
}
else {
cout << "There are " << words_in_string << " words in the string." << '\n';
}
return (0);
}
/*****************************************************************
* word_count -- counts words in a given string. Considers a period, comma and space *
* bar as word seperators. Uses ASCII code to determine. *
* *
* Parameters *
* some string *
* *
* Returns *
* num_words -- number of words in string *
******************************************************************/
int word_count(char input_string) // funtion to find how many words
{
int counter = 0;
char variable_x;
int num_words = 0;

for (variable_x = input_string[0]; variable_x != '\0'; counter++) { // looping through string
if((variable_x == 32) || (variable_x == 46) || (variable_x == 44)) { // looking for spaces
num_words++;
}
variable_x = input_string[counter];
}
return (num_words); // answer to how many words.
}

// my compiler errors were: (c++ source file was called 9-1.cc Both errors within the function)
// the first error was the beginning of the for loop, the second was "variable_x = input_string[counter]; " Can I not do this?

//9-1.cc:38: error: invalid types `char[int]' for array subscript
//9-1.cc:42: error: invalid types `char[int]' for array subscript

Rico16135 · 04-26-2004, 04:52 PM

forgive me, I think the word wrap screwed up some of the code. Especially the "{}" Excuse the code. Just look over the concepts if you could. Thanx guys.

thallium205 · 05-04-2009, 10:21 AM

Code:

#include <iostream>
#include <string>
using namespace std;

int main()
{
	int i, numspaces;
	char nextChar;
	string msg;

	numspaces=1;

	cout << "Type in a string\n";
	getline(cin, msg);

	// checks each character in the string
	for (i=0; i<int(msg.length()); i++)
	{
		nextChar = msg.at(i); // gets a character
		if (isspace(msg[i]))
			numspaces++;
	}
	cout << "\nThere are " << numspaces << " words in this string.";
	cin.ignore();
	return 0;
}

rriggs · 05-04-2009, 05:07 PM

Iterators make quick work of the problem.

Code:

#include <iostream>
#include <sstream>
#include <string>
#include <iterator>

using namespace std;

int main()
{
	string msg;

	cout << "Type in a string\n";
	getline(cin, msg);
	
	istringstream ostr(msg);
	istream_iterator<string> it(ostr);
	istream_iterator<string> end;
	
	size_t words = 0;
	while (it++ != end) words++;

	cout << "\nThere are " << words << " words in this string." << endl;

	return 0;
}

amysaraantony · 05-06-2009, 11:56 AM

Googling for this should get you all the code you want my friend.

Debian

dmail · 05-06-2009, 12:30 PM

Please stop keeping this thread alive it was started over five years ago and your comment did not add anything of value. In fact from looking at your post history, I do not think any of your posts have added value