Write a program that translates an integer consisting of upto 4 binary digits into its binary equivalent.

Input:
• xxxx where each x is either a 0 or 1
• there could be 1, 2, 3, or 4 digits

Processing:
• Use remainders to strip off each of the possibly 4 digits
• It really doesn't matter if there are 1, 2, 3, or 4 digits since the remainder would just be 0
• Using the example of 1010, divide by 1000. The integer result is a 1. This is your left most digit
• If you take the remainder of 1010 after dividing by 1000, then the answer is 010 or 10.
• Divide 10 by 100. The integer result is a 0. This is your second digit
• Take the remainder of 010 after dividing by 100, then the answer is 10.
  
• Do you see a pattern evolving?

Output:
• the decimal equivalent of the binary number xxxx
• For example 0011 would result in a 3 being displayed

How to make this work
• The secret is in the division AND being able to change the original number.
• In programming, when you divide an integer by another integer, the decimal part is truncated. Consider this code snipet:
  int binary_string;
  int digit4;
  digit4 = binary_string/1000;
  
  Since this is integer division the 010 of 1010 will be truncated and digit4 will be a one.
• Now, how to extract the remainder after division... We could use a mathematical formula (originalNumber - digit4*1000), or we could use C++ builtin operator for modulus(remainder). The % sign is this operator.
  binary_string % 1000 will result in 010 the remainder after division.
• We want the new binary_string that we are going to work with to be changed from 1010 to 010.
  binary_string = binary_string % 1000;
  Code Output
• Now repeat for the other digits
  Code Output

Now that the binary number is split apart, how do we change the digits into the decimal equivalent?
Notice that digit4 is the 8's place holder, digit 3 is the 4's place holder, digit2 is the 2's place holder and the result in binary_string is the 1's place holder. So, we could use:
digit4*8 + digit3*4 + digit2*2 + binary_string
Here's the code Here's the Output

OR

digit4*(2*2*2) + digit3*(2*2) + digit2*(2) + binary_string
Notice that digit4 is multiplied by three 2s, digit3 by two 2s,... Here's a neater solution and the output

Notice that the binary string that is printed at the end of the code is incorrect. We need to print it at the beginning instead.
solution and output