C++ Program to Convert Binary Code to Gray Code

This is a C++ Program to Convert a Number in Binary Representation to Gray Code.

Problem Description

The program takes a binary number and converts it into its equivalent Gray code. In Gray code, any two successive values differ only by one bit.

Problem Solution

1. A binary number is entered.
2. Using a while loop, the binary number is converted to its equivalent Gray code, using XOR operation.
3. The result is printed.
4. Exit.

C++ Program/Source code

Here is the source code of C++ Program to Convert a Number in Binary Representation to Gray Code. The program output is shown below.

```
#include <iostream>
```
```
#include <math.h>
```
```
using namespace std;
```
```
int main ()
```
```
{
```
```
    int num, temp, x, y, i = 0, bin =0;
```
```
    cout << "Enter a binary number : ";
```
```
    cin >> num;
```
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	temp = num;
```
```
    while (temp != 0)
```
```
    {
```
```
        x = temp % 10;
```
```
        temp = temp / 10;
```
```
        y = temp % 10;
```
```
        if ((x && !y) || (!x && y))
```
```
            bin = bin + pow(10, i);
```
```
        i++;
```
```
    }
```

    cout << "\nGray code of " << num << " is : " << bin;

```
    return 0;
```
```
}
```

Program Explanation

1. The user is asked to enter a binary number and it is stored in the variable ‘num’.
2. num is copied in a temporary variable ‘temp’. ‘bin’ and ‘i’ are initialized as 0.
3. Using a while loop, temp is divided by 10 and the remainder is stored in ‘x’.
4. temp is divided again and the remainder is stored in ‘y’.
5. XOR of x and y is calculated. If it is true, 10 to the power i is added to bin.
6. i is incremented.
7. The loop continues till temp is equal to 0.
8. bin is then printed which is the Gray code equivalent.

Runtime Test Cases

Case 1 :
Enter a binary number : 011
Gray code of 011 is : 10
 
Case 2 :
Enter a binary number : 10101010
Gray code of 10101010 is : 11111111
 
Case 1 :
Enter a binary number : 1110
Gray code of 1110 is : 1001

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