Encrypt Digit C++ - c++

Question

So I am trying to encrypt a four digit integer by adding seven to the digit then dividing the whole digit by ten. In my program I am taking each single digit separately and then I need to divide the whole digit by ten. How can I combine all the separate int into one four digit number?
#include "stdafx.h"
using namespace std;
int main()
{
//Define all variables needed
int a,b,c,d,enc,ext;
//Print dialog and input each single digit for the four number digit
cout << "Enter Your First Digit:" << endl;
cin >> a;
cout << "Enter Your Second Digit:" << endl;
cin >> b;
cout << "Enter Your Third Digit:" << endl;
cin >> c;
cout << "Enter Your Fourth Digit:" << endl;
cin >> d;
//Add seven to each digit
a += 7;
b += 7;
c += 7;
d += 7;
a /= 10;
b /= 10;
c /= 10;
d /= 10;
cout << "Your encrpyted digits:" << c << d << a << b <<endl;
cout << "Enter 1 to exit:" <<endl;
cin >> ext;
if (ext = 1) {
exit(EXIT_SUCCESS);
}
}
As you probably noticed I am dividing each number separately. I need to do them together. Then I am also creating a decrypting which I will get me back to the original number in a separate program.

Answer 1

Based on your comment you are trying to do a variation on the Caesar Cipher, in which case you should be using the modulus operator (%) not the integer division operator (/). Using integer division loses information which will prevent you from decrypting. When your digit is in {0, 1, 2} your division results in a 0. When it is in {3, 4, 5, 6, 7, 8, 9}, the division results in a 1. You can't decrypt {0, 1} back into the original number without some additional information (which you have discarded).
If you want to encrypt on a digit by digit basis using the Caesar Cipher approach, you should be using modulo arithmetic so that each digit has a unique encrypted value which can be retrieved during decryption. If that's really what you are looking for then you should be doing something like the following to encrypt with a 7:
a = (a + 7) % 10;
b = (b + 7) % 10;
c = (c + 7) % 10;
d = (d + 7) % 10;
To decrpyt, you subtract 7, which in mod 10 arithmetic is an addition by 3, so that would be:
a = (a + 3) % 10;
b = (b + 3) % 10;
c = (c + 3) % 10;
d = (d + 3) % 10;
This of course presupposes you've properly validated your input (which isn't the case in your example above).

Answer 2

Combining the individual digits into one four-digit number is simple; just multiple the first digit by 1000, add the second multiplied by 100, and so on.
But this is a one-way algorithm; you will never be able to retrieve the original four-digit number from this.

Answer 3

This is what youd'd probably be looking for :
int e = (a*1000)+(b*100)+(c*10)+d;
e=e/10;

Answer 4

It's not clear from your description whether the addition should be modulo 10 or not; if so
((((((a % 10) * 10) + (b % 10)) * 10) + (c % 10)) * 10) + (d % 10)
if you don't want the modulo 10
(((((a * 10) + b) * 10) + c) * 10) + d

Answer 5

Stepping aside the fact that you almost certainly want mod instead of divide (as #Andand has said), there's more than one way to turn the digits into a number!
A lot of people using interpreted languages these days would probably want to do it symbolically. C++ can do that too, fairly neatly in fact:
// create a string stream that you can write to, just like
// writing to cout, except the results will be stored
// in a string
stringstream ss (stringstream::in | stringstream::out);
// write the digits to the string stream
ss << a << b << c << d;
cout << "The value stored as a string is " << ss.str() << endl;
// you can also read from a string stream like you're reading
// from cin. in this case we are reading the integer value
// that we just symbolically stored as a character string
int value;
ss >> value;
cout << "The value stored as an integer is " << value << endl;
It won't be as efficient as multiplications in this narrow case of a 4 digit number, because of the round trip to a string and back. But good to know the technique. Also it's a style of coding that can be a lot easier to maintain and adapt.
You'll get stringstream if you #include <sstream>.