y or n option?

Could some one give me a yes or no example. But with y and n. Also could you add something to make it where case doesn't matter (if you want to)? Something like-

Code:

---

cout << "Again?: ";
cin >> again;
if again = "y"
    cout << "\nYou said yes.";
if again = "n"
    cout << "\nYou said no.";
else
    cout << "\nNot a valid option.";

---

I appologize as I do not know the usage of a if statement in C++.

Code:

---

#include <iostream>
#include <cctype>

using std::cin;
using std::cout;

int main()
{
  char yn;

  cout<<"Again? (y/n): ";
  cin>> yn;

  yn = tolower ( yn );

  if ( yn == 'y' )
    cout<<"You said yes!"<<std::endl;
  else if ( yn == 'n' )
    cout<<"Party pooper :("<<std::endl;
  else
    cout<<"I said \'y\' or \'n\' silly."<<std::endl;
}

---

-Prelude

Thank you so much for that, but I have two questions. What is tolower? And in your if statement, why do you use std::endl instead of just using std::endl at the top?

>What is tolower?
It's a standard library function that converts a char to its lower case equivalent or does nothing if there is no lower case equivalent. It's located in the cctype header.

>why do you use std::endl instead of just using std::endl at the top?
I forgot, adding the using statements was a last minute decision. ;)

-Prelude

tolower returns the letter argument in lowercase, regardless of it's case going in.

She does this because ('y' == 'Y') is false in C++.

As for not adding a using directive for std::endl, I can only tell you that it must be correct, since prelude is perfect, but I don't have a clue as to her reasoning.

for std stuff, you could do any of the following

Code:

---

using namespace std;
OR
using std::endl;
OR
//like what prelude did
std::cout << std::endl;

---

all of them work, it's up to the user as to how he or she wants to do it.

I now learned about the tolower fxn...I probably would have done a switch, like this:

Code:

---

switch(yn) {
    case 'y':
    case 'Y':
        cout << "You said yes" << endl;
        break;
    case 'n':
    case 'N':
        cout << "You said no" << endl;
        break;
    default:
        cout << "Invalid letter." << endl;
        break;
}

---

but the tolower is easier.

>all of them work, it's up to the user as to how he or she wants to do it.
There are distinct differences though. I prefer explicit std:: resolution most of the time because the notational convenience of using statements really isn't worth cluttering the global namespace. I never use

using namespace std;

because it's like saying "I'll only use cout and endl, but give me everything just in case". That's poor practice even though I know better than to use a standard name for something I write.

-Prelude

Quote:

---

Originally posted by Imperito
She does this because ('y' == 'Y') is false in C++.

---

Yea, you could have just said C++ is case sensitive. Or if you looked at my post and see where I said

Quote:

---

add something to make it where case doesn't matter

---

and that gives you a little clue that i already knew C++ is case sensitive. But thanks for that anyways. ;)

well ii use something like this.. as long as the user enters 'y as the option the program keeps running;;;;

Code:

---


# include <iostream.h>

int main()
{
char option='y';

while(option=='y')
      {

          // the program you want to run....
         
          cout<<"Do you want to run the prog again (y/n) > ";
          cin>>option;

if(option!=n && option!=y)
option='y';

      }


return 0;
}

---

Why not step it up a bit:

Code:

---

#include <iostream>
#include <conio>
#include <cctype>
int main()
{
 cout<<Again (y/n): ";
 char yn = getch();
 yn = tolower ( yn );
 if(yn == 'n')
    cout<<"Loser";
 if(yn == 'y')
    cout<<"DUDE!!!";
 else
    return 0;
}

---

Hi,

getch() is a non-standard function.... hence its preferred that you don't use getch() in your programs.. unless you explicitly work with Turbo C / C++ and are stuck to DOS or windows platform.

Quote:

---

Originally posted by devour89
Why not step it up a bit:

Code:

---

#include <iostream>
#include <conio>
#include <cctype>
int main()
{
 cout<<Again (y/n): ";
 char yn = getch();
 yn = tolower ( yn );
 if(yn == 'n')
    cout<<"Loser";
 if(yn == 'y')
    cout<<"DUDE!!!";
 else
    return 0;
}

---

---

On line 6 you have no opening quotation mark. Just thought I would say something.