[ishkabible]

start by telling us what you are trying to do then tell us what as happening as opposed to what you think should happen. then we may be able to help

[shadon]

plz could any one help !!

[TomJoad]

I've been using learncpp.com to, well, learn c++. Anyways, section 6.8 (http://www.learncpp.com/cpp-tutorial/68-pointers-arrays-and-pointer-arithmetic/)has almost exactly what you are looking for in the last snippet of code.

Now if I could only figure out a way so it will count words instead of letters...

[#define]

Choice 1 works reasonably well,if you remove break.
Added int before main(), and braces for choice A if() statement.

Choice 2 is not working, y is already integer variable, the for statement continuously reads cin to y, and you can't put q in to exit.
Didn't look any further.

Look at using functions to break up your code.

#include<iostream.h>

//int added
int main()
{

  int i, y, rc=0, ec=0, ac=0, n=0;

  char choice;
  char x;

  cout << "please enter your choice\n";
  cout << "press a or A to count characters\n" 
       << "press b or B to sinx and series\n"
       << "press c or C to count perfects,digits and prime number\n" 
       << "press d or D to Binary and Decimal\n" 
       << "press d or D to exit\n";

  cin>>choice;

  if(choice=='a'||choice=='A')
  { // braces added
    cout << "press 1 to count characters\n" 
         << "press 2 to change text number to integer\n" 
         << "press 3 to change from capital to small\n" 
         << "press 4 to return to the main menu\n";

    cin>>choice;

    if(choice=='1')
    {
      cout<<"enter your text\n";
       
      for(;x!='q';)/>
      {
        cin>>x;
        if(x=='a')
          ac=ac+1;

        else if(x=='r')
          rc=rc+1;

        else if(x=='e')
         ec=ec+1;

        cout <<  "a=" << ac
             << " r=" << rc
             << " e=" << ec << endl;
        // break;
      }
    }
   

    if(choice=='2')
    {
      cout<<"enter your text number\n";

      // for(;y!='q';)/>
      // y is not character
      cin >> y;

      if(y>=48 && y<=57)
        n=n*10+(y-48);

      cout<<"The int is " << n << "\n";
    }


  }

  cout << "Press any key to exit";
  cin.ignore();
  cin.get();
  return(0);
}

[shadon]

TomJoad

thank you for your help i really appreciated it

define

thank you i will try it , i cant find the words to thank you so