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Answer to Question #22176 in C++ for rodea embradura

Question #22176
create a c++ program that performs multiplication and division of binary and hexadecimal.

sample output:

number system

1. binary
2. hexadecimal
enter your choice: 1

enter first binary number: 1011
enter second binary number: 11

arithmetic operation
1. multiplication
2. division
enter your choice: 1

result: 100001
Expert's answer

#include <iostream>
#include <bitset>
#include <limits>

using namespace std;

/* mutiplication using the binary shift operator */
int multiplication(int mul, int by)
{
int result = 0;
if (by > 0)
{
// mulitiply by (mulby) comparison for when to add a shifted value to the
result.
// moveby, how many shifts left to shift by if there is a match.
int mulby = 2, moveby = 1;
while (by >= mulby)
{
cout << "by = " <<
bitset<numeric_limits<short>::digits>(by) << endl;
cout << "mulby = " <<
bitset<numeric_limits<short>::digits>(mulby) << endl;
// bitwise add and if it is the same value as the mulby then add to the result.
if ((by & mulby) == mulby) {
cout << "+ result= " <<
bitset<numeric_limits<short>::digits>((mul << moveby))
<< endl;
result += mul << moveby;
cout << "result = " <<
bitset<numeric_limits<short>::digits>(result) << endl;
}
mulby = mulby << 1; // the next test if the bits values are the same.
moveby++; // move the result by
}
// and if there is a modulus of 1 from 2 then the value was odd.
if (by % 2 == 1)
{
result += mul;
}
cout << "end result = " <<
bitset<numeric_limits<short>::digits>(result) << endl;
}
return result;
}

int main ()
{
cout << multiplication(6,7) << endl;
return 0;
}

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