Input an integer containing only 0 s and 1 s (i.e., a "binary" integer) and
print its decimal equivalent. Use the modulus and division operators to pick
off the "binary" number's digits one at a time from right to left. Much as in
the decimal number system, where the rightmost digit has a positional value of
1, the next digit left has a positional value of \(10,\) then \(100,\) then
\(1000,\) and so on, in the binary number system the rightmost digit has a
positional value of \(1,\) the next digit left has a positional value of \(2,\)
then \(4,\) then \(8,\) and so on. Thus the decimal number 234 can be interpreted
as \(2 * 100+3 * 10\) \(+4 * 1 .\) The decimal equivalent of binary 1101 is \(1 *
1+0 * 2+1 * 4+1 *\) 8 or \(1+0+4+8,\) or \(13 .\) [Note: The reader not familiar
with binary numbers might wish to refer to Appendix D.]