Add, Subtract, Multiplication, Division

How do you implement a function to add two integers without utilization of arithmetic +, -, *, and / operators?
Bit operations The first step is to add digits without carries. The results are 0 when adding 0 and 0, as well as adding 1 and 1. The results are 1 when adding 0 and 1, as well as 1 and 0. Note that these results are the same as bitwise XOR operations. The bitwise XOR results of 0 and 0, as well as 1 and 1, are 0, while the XOR results of 0 and 1, as well as 1 and 0, are 1. The second step is for carries. There are no carries while adding 0 and 0, 0 and 1, as well as 1 and 0. There is a carry only when 1 and 1 are added. These results are the same as bitwise AND operations. Additionally, we have to shift carries to the left for one bit to get the actual carry value. The third step is to add results of the first two steps. The adding operations can be replaced with bit operations again. These two steps above are repeated until there are carries. int add(int num1, int num2) {
    int sum, carry;
    do {
        sum = num1 ^ num2;
        carry = (num1 & num2) << 1;
        num1 = sum;
        num2 = carry;
    } while (num2 != 0);

    return num1;
}
How do you implement a function for the subtraction operation without utilization of arithmetic +, -, *, and / operators?
-b can be gotten with bit operations because -b=~b+1.
int subtract(int num1, int num2) {
    num2 = add(~num2, 1);
    return add(num1, num2);
}
How do you implement a function for the multiplication operation without utilization of arithmetic +, -, *, and / operators?
a×n can be implemented with a left-shift and additions. Since there are O(logn) 1 bits in the binary representation of n, the new solution invokes the add method for O(logn) times.
int multiply(int num1, int num2) {
    boolean minus = false;
    if ((num1 < 0 && num2 > 0) || (num1 > 0 && num2 < 0))
        minus = true;

    if (num1 < 0)
        num1 = add(~num1, 1);
    if (num2 < 0)
        num2 = add(~num2, 1);

    int result = 0;
    while (num1 > 0) {
        if ((num1 & 0x1) != 0) {
            result = add(result, num2);
        }

        num2 = num2 << 1;
        num1 = num1 >> 1;
    }

    if (minus)
        result = add(~result, 1);

    return result;
}
How do you implement a function to divide an integer by another without utilization of arithmetic +, -, *, and / operators?

Array Construction
void multiply(double array1[], double array2[]){
    if(array1.length == array2.length && array1.length > 0){
        array2[0] = 1;
        for(int i = 1; i < array1.length; ++i){
            array2[i] = array2[i - 1] * array1[i - 1];
        }

        int temp = 1;
        for(int i = array1.length - 2; i >= 0; --i){
            temp *= array1[i + 1];
            array2[i] *= temp;
        }
    }
}
The time complexity of this solution is O(n) obviously