Counting Words

[Alicito]

well my friends I tried to solve a problem and it was about allowing the user to enter a paragraph , then the compiler'll count and display the number of typed words .

welll I easily solved it like this

--------------------------------------------------

#include<iostream>
#include<string>
using namespace std;
void main()
{
	int words=0;
	string str;
	getline(cin,str);
	for(int i=0;i<(int)str.length();i++)
		if(isspace(str[i]))
			words++;
	cout<<"number of typed words is: "<<words+1<<endl;
}

-----------------------------------------------------------------------------------
but later I relized that it was very weak program
coz some users use tabs just move the pointing crosser
things like these
so who has a better idea ?

[mvmalderen]

Would you mind reading this and use code tags ?

Edit:: And something else: migrate your code to standard C++ code (don't use [B]void main[/B] for example), review this web page carefully :)

Edit:: Tom Gunn's (one post below, #3) is right in saying that it will check for all space characters, look at this

[Tom Gunn]

For a loose definition of 'word', you can use cin:

string word;

while (cin >> word)
{
    cout << word << '\n';
}

For a stronger definition of 'word', you should do more parsing than just looking for spaces. Punctuation and stuff like that wouldn't be a part of the word, right?

edit
By the way, isspace checks for all space characters: {' ', '\t', '\r', '\n', '\v', '\f'}. Your program should work with tabs too, but you don't take multiple space characters between words into account, so the word count is wrong.

[mvmalderen]

The following:

for(int i=0;i<(int)str.length();i++)
if(isspace(str[i]))
words++;

has to be:

for(int i=0;i<(int)str.length()[B]-1[/B];i++)
    if(isspace(str[i]))
        words++;

Edit:: No, this won't work either

[Ancient Dragon]

you could use stringstream class in <sstream>

string sentence = "Now is the time for all old folks to be in bed";
string word;
stringstream str(sentence);
while( str >> word)
   cout << word << "\n";

[siddhant3s]

Why not use stringstreams?

#include<iostream>
#include<string>
#include<sstream>
int main()
{
    std::string str;
    getline(std::cin,str);
    std::stringstream sstrm(str);
    int count=0;
    std::string temp;
    while(sstrm>>temp) count++;
    
    std::cout<<count<<std::enl;
}

output:

siddhant3s@Xion:~$ g++ tester.cpp -o tester.exe -ansi -pedantic && ./tester.exe
Let me type something   which 	also contains tabs
8
siddhant3s@Xion:~$

Edit: [siddhant3s wants to question the world why he every time late by few seconds. @Ancient Dragon]

[Tom Gunn]

Why not use stringstreams?

They add unnecessary complexity?

#include <iostream>
#include <string>

using namespace std;

int main()
{
    string str;

    while (cin >> str) count++;
    
    cout << count << '\n';
}

[Ancient Dragon]

They add unnecessary complexity?

And, with your otherwise excellent solution, you have to terminate input with Ctrl+Z instead of just <Enter>

[mvmalderen]

They add unnecessary complexity?

#include <iostream>
#include <string>

using namespace std;

int main()
{
    string str;

    while (cin >> str) count++;
    
    cout << count << '\n';
}

The word counting will never stop... (unless you terminate your program in an improper way)

Edit:: AD was before me :P

[siddhant3s]

>They add unnecessary complexity?
Oh stop kidding......
They are simplifying the program not making it more complex. Your example won't work if you intend to write a function to count the number of words in a given string.

[Tom Gunn]

And, with your otherwise excellent solution, you have to terminate input with Ctrl+Z instead of just <Enter>

The problem was to allow the user to enter a paragraph. I bet there will be more than one line, and to handle multiple lines with getline you need to terminate input with something besides <Enter>.

>They add unnecessary complexity?
Oh stop kidding......
They are simplifying the program not making it more complex. Your example won't work if you intend to write a function to count the number of words in a given string.

Who's kidding? You're adding more complexity by making the code 'function ready' when Alicito's code does everything in main. Besides, if you want to write a function you can pass an istream and use the same short code without hurting flexiblity. It works with stringstreams and fstreams too:

#include <iostream>
#include <string>

using namespace std;

unsigned long CountWords(istream& stream)
{
    unsigned long count = 0;
    string str;

    while (stream >> str) ++count;

    return count;
}

int main()
{
    cout << CountWords(cin) << '\n';
}

Ancient Dragon's complaint makes more sense, but only if you're reading one line interactively.

[mvmalderen]

The problem was to allow the user to enter a paragraph. I bet there will be more than one line, and to handle multiple lines with getline you need to terminate input with something besides <Enter>.

Hey great programmer, I bet you forgot something :P
Your program doesn't ignore punctuation marks, a great programmer as you would certainly have seen that, not?

e.g.:
Are these: "Hello , World !!!" four words or two ?

[Ancient Dragon]

>>four words or two
Two. Punctuation marks are not considered words. None of the solutions presented so far in this thread would parse that sentence correctly.

[Tom Gunn]

Your program doesn't ignore punctuation marks, a great programmer as you would certainly have seen that, not?

I'm not a great programmer by any measure, but I did say that the definition of 'word' is important in my first post:

For a loose definition of 'word', you can use cin:

string word;

while (cin >> word)
{
      cout << word << '\n';
}

For a stronger definition of 'word', you should do more parsing than just looking for spaces. Punctuation and stuff like that wouldn't be a part of the word, right?

Salem jumped me about giving away code, so I didn't give an example of the stronger definition. Blame him. ;)

[mvmalderen]

>Salem jumped me about giving away code, so I didn't give an example of the stronger definition. Blame him.
No, giving a stronger definition about the word 'word' is not the same as giving away free code :)