Counting characters, words, spaces, punctuations, etc.

Applications Programming
1

I am very new to C programming.

How could I write a C program that could count the characters, words, spaces, and punctuations in a text file?

Any help will be really appreciated. I am doing this as part of my C learning exercise.

Thanks,

Ajay

2

I hope this is not homework.

#include <ctype.h> defines the functions and macros you need.
isspace() spaces (include newlines)
ispunct() punctuation characters
isalnum() regular characters A-z a-z plus digits 0-9

  1. open the file with fopen

2 use fgets to read a line

while you get a line of text
3. interate over all of the chacters you just read in - checking char types with those
functions
end while

  1. close the file

  2. use printf to display all of the results. word count per line == # of isspace() characters you find on the line, so it also equals the total words. This assumes only single spaces between characters. And no leading spaces on a line.

3

If this is not homework then post here what you have tried to code so far...

4

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

int main () {
string lineBuffer;
ifstream inMyStream ("c:/File.txt");
if (inMyStream.is_open()) {
int count = 1;
while (!inMyStream.eof() ){
getline (inMyStream, lineBuffer);
cout << count << lineBuffer << endl;
count++;
}
inMyStream.close();
}
else cout << "File Error: Open Failed";
return 0;
}

but, that is where i got stuck.

5

/use this to count words/
#include <stdio.h>
#define IN 1 /* inside a word*/
#define OUT /outside a word/

int
main(void)
{
int nw, state;
state = OUT;
while((c= getchar()) != EOF){

 /*think using something here for punctuations*/
 /*for char count*/ 

  if\(c == ' ' || c == '\\n' || c == '\\t'\)
        state = OUT;
   else if \(state == OUT\)\{
        state = IN;
        \+\+nw;  
   \}

}
return 0;
}

/see it and try to implement otherway/