C++ : WordCount from Char string with pointers....?

Question

I want to pass a char string as a pointer reference and then count the words in this string ... but somehow I never get to count the right number of words... here my code:

#include <iostream>
#include <stdio.h>
using namespace std;

int charCount(const char* pPtr);

int main() {

    char wort[] = "Ein Neger mit Gazelle zagt im Regen nie ";
    int count(0);

    count = charCount(wort);
    cout <<count <<endl;

}

int charCount(const char* pPtr) {
    int wordCount(0);

    while(*pPtr != '\0') {      

        //Falls EOF Erreicht und vorheriger Buchstabe war kein Blank oder newline dann Wortzaehler erhoehen
        if ((*pPtr == '\0') && (*(pPtr-1) !=' ' || *(pPtr-1) != '\n')) {
            wordCount++;

        }

        //Falls Blank oder Newline, und vorheriger Buchstabe war kein Blank oder Newline, Wortzaehler erhoehen
        if (((*(pPtr+1) == ' ' || *(pPtr+1) == '\n')) && ((*(pPtr) != ' ' || *(pPtr) != '\n' ))) {
                wordCount++;                
        }       
        pPtr++;

    }
    return wordCount;
}

I got 7, but when I add blanks at the end, he counts them as words... — MarcL
– MarcL, Commented Nov 15, 2013 at 17:36

ordahan · Accepted Answer · 2013-11-15 17:21:20Z

1

Looks like the while(*pPtr != EOF) should actually be while(*pPtr != NULL).

EOF on some systems is 0 (like NULL) and on some it might be -1 or any other value.

Also, looks like a better approach to your problem is having some kind of 'state-machine', i.e:

int in_word = 0;
while (*pPtr != NULL){
    if ((*pPtr >= 'a' && *pPtr <= 'z') || <same for uppercase>){
        in_word = 1;
    }
    else if (in_word == 1){
        wordCount++;
        in_word = 0;
    }

Not sure if this covers everything.. but I hope you get the general idea.

edited Nov 15, 2013 at 17:21

answered Nov 15, 2013 at 17:15

ordahan

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4 Comments

twalberg Over a year ago

I don't believe I've ever seen a system where EOF == 0. I will admit, though, that I have not perused every available system...

ordahan Over a year ago

I believe that in anything ANSI / Modern etc you will in fact never find it. Here you go: stackoverflow.com/a/1624150/1778249

MarcL Over a year ago

this is not very helpful, it doesnt even execute correctly... has nobody an idea how to realise a wordcount with pointers please?

MarcL Over a year ago

EDIT: Actually , it was very helpful ;) How can I expand it for alphanumeric characters , also with numbers and other characters, just not for blanks and newline characters please?

Hernán Tellechea · Accepted Answer · 2013-11-15 18:01:59Z

1

I think the easiest way to count words in a string is using stringstreamming!

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

int charCount(const char* pPtr);

int main() {
    char wort[] = "Ein Neger mit Gazelle zagt im Regen nie ";
    int count(0);
    count = charCount(wort);
    cout <<count <<endl;
}

int charCount(const char* pPtr) {
    int wordCount(0);
    stringstream ss;
    string temp;
    ss<<string(pPtr);
    while(ss>>temp)
        wordCount++;
    return wordCount;
}

http://www.cplusplus.com/reference/sstream/stringstream/

answered Nov 15, 2013 at 18:01

Hernán Tellechea

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Comments

JoeC · Accepted Answer · 2013-11-15 18:19:02Z

1

I think it should be while(*pPtr != '\0')

Note that '\0' is the end of a char array, and generally, EOF is -1, but there is no -1 in your char array so the loop will go above your char array until it finds -1

int charCount(const char* pPtr) {
int wordCount(0);
int track;
while(*pPtr != NULL) {       
    if ((*pPtr == ' ' || *pPtr == '\n' || *pPtr == '\r')&& track != 0){
        //cout << *pPtr << endl;
        wordCount++;
        track = 0;
    }else if ((*pPtr != ' ' && *pPtr != '\n' && *pPtr != '\r')){
        track++;
    }
    pPtr++;
}
return wordCount;
}

Try this!

edited Nov 15, 2013 at 18:19

answered Nov 15, 2013 at 17:19

JoeC

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1 Comment

MarcL Over a year ago

ok I added '\0' but I had this before and he still counts wrong , blanks at the end = words...

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