Well I'm trying to create this one game for my school project but it is supposed to be a "magic" trick a but it will defeat the purpose if you can see the number in the screen so I was wondering if you can amke to not show the input a and c.

#include <iostream>
#include <string>

using namespace std;
int main () {
int a, b, c;
string sg;
sg = "First number is the one you chose the rest are your age.";
cout << "MAGIC Numbers. \n";
cout << "Pick a number from one though ten:";
cin >> a;
a = a * 2;
a = a + 5;
a = a * 50;
cout << "Did your birthday already pass? \n (Enter 1 for yes 2 for no) \n";
cin >> b;
if ( b == 1 ){
a = a + 1761;
}
else {
a = a +1760;
}
cout << "Enter your year of birth: \n";
cin >> c;
a = a - c;
cout << sg << a <<"\n";
system ("pause");
}

You can use getch() to input single character, which will not be displayed over screen.
ch = getch();

And, user don;t need to press enter key here.



Gorav
http://www.kgsepg.com

Thanks

I would use cin.get() instead of getch().

How would you use cin.get() im trying it but cant figure it.

#include <iostream>
#include <string>

using namespace std;
int main () {
int a, b, c;
string sg;

sg = "First number is the one you chose the rest are your age.";
cout << "MAGIC Numbers. \n";
cout << "Pick a number from one though ten:";
cin.get (a); // Here?
a = a * 2;
a = a + 5;
a = a * 50;
cout << "Did your birthday already pass? \n (Enter 1 for yes 2 for no) \n";
cin.get >> b;
if ( b == 1 ){
a = a + 1761;
}
else {
a = a +1760;
}
cout << "Enter your year of birth: \n";
cin.get >> c;
a = a - c;
cout << sg << a <<"\n";
system ("pause");
}

Either:

cin.get(c);
or
c = cin.get();

Note that doing it this way will *not* remove the enters that will be in the buffer from when the user sends you input. You will have to deal with those yourself somehow (probably using cin.ignore()).