Program to count "the" word?

[princemoon]

I made this program to count vowels in given words now what should I change in it to count word only 'the' in this program?

#include <iostream.h>
#include<conio.h>

int main()
{
char ar[20];
int vowc=0,i;
cout<<("Enter the array\n");
gets(ar);
for(i=0;i<20;i++)
{
switch(ar[i])
{
case 'a' :
case 'A' :
case 'e' :
case 'E' :
case 'i' :
case 'I' :
case 'o' :
case 'O' :
case 'u' :
case 'U' :
vowc++;
break;
default : break;
}
}
cout<<("%d",vowc);
return 0;
}

[vlad from moscow]

Use C function strstr or you can change the program the following way

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
   char ar[20];
   unsigned int found = 0;
   unsigned int count = 0;

   cout << "Enter the array\n";
   gets( ar );

   for ( char *p = ar; *p; ++p )
   {
      switch ( *p )
      {
         case 't' :  case 'T' :
            found = 1;
            break;
         case 'h' : case 'H' :
            if ( found ) ++found;
            break;
         case 'e' : case 'E':
            if ( found == 2 ) count++;
            found = 0;
            break;
         default:
            found = 0;
            break;
      }
   }

   cout << count;

   return 0;
}

EDIT: Also I think you do not understand what happens in the statement of your original program

cout<<("%d",vowc);

Though this statement works however it has no any sense that is there is no any sense to use the comma operator. You should write simply

cout << vowc;

[princemoon]

Thanx alot dear you did it right & I will keep in mind the comma use next time.

[vlad from moscow]

However the suggested code contains a bug! :) Shall be

case 'h': case 'H':
	if ( found == 1 ) ++found;
	else found = 0;
	break;
case 'e': case 'E':
	if ( found == 2 ) count++;
	found = 0;
	break;

[vlad from moscow]

Also instead of gets it is better to use cin.getline( ar, sizeof( ar ) );