solve divide two integers problem using bit manipulation

Bit-manipulationTechniques/DivideTwoIntegers/main.cpp

@@ -0,0 +1,63 @@
+#include <iostream>
+#include <climits>
+
+using namespace std;
+
+int divide(int dividend, int divisor)
+{
+    if (dividend == divisor)
+    {
+        return 1;
+    }
+
+    bool isPositive = dividend > 0 == divisor > 0; // true both dividend and divisor have the same sign
+    unsigned int divd = abs(dividend);             // short for dividend since this name is already used
+    unsigned int divs = abs(divisor);              // short for divisor since this name is already used
+    unsigned int ans = 0;
+
+    // dividend = quotient * divisor + remain
+    while (divd >= divs)
+    {
+        int exp = 0;
+
+        /*
+         * "<<" left-shift bitwise operator. a << b moves the bits in "a" by "b"
+         * positions. Given as a result: a * 2^b
+         *
+         * This loop will increment the exp until the result of:
+         *   divisor * 2^exp
+         * is greater than the dividend.
+         */
+        while ((divs << (exp + 1)) < divd)
+        {
+            exp++;
+        }
+
+        /*
+         * This will end up in adding all the powers of 2 that compose the quotient
+         * i.e. getting the value of quotient by a binary representation.
+         */
+        ans += (1 << exp); // ans += 2^exp
+        divd = divd - (divs << exp);
+    }
+
+    if (ans == (1 << 31) && isPositive)
+    {
+        return INT_MAX;
+    }
+
+    return isPositive ? ans : -ans;
+}
+
+int main()
+{
+    int dividend;
+    cin >> dividend;
+
+    int divisor;
+    cin >> divisor;
+
+    cout << "Result: " << divide(dividend, divisor) << endl;
+
+    return 0;
+}