Add, Subtract Multiply, Division without arithmetic operator - Stack Overflow

写一个函数，求两个整数之和，要求在函数体内不得使用+、-、x、/四则运算符号。
看来只能使用位运算了。

private static int add(int num1, int num2){

    int sum = 0;

    int carry = 0;

    do{

        sum = num1 ^ num2; // no need use sum

        carry = (num1 & num2)<<1;

        num1 = sum;

        num2 = carry;

    }while(num2 != 0);

     

    return num1;

}

int Add(int x, int y)

{

    // Iterate till there is no carry 

    while (y != 0)

    {

        // carry now contains common set bits of x and y

        int carry = x & y; 

        // Sum of bits of x and y where at least one of the bits is not set

        x = x ^ y;

        // Carry is shifted by one so that adding it to x gives the required sum

        y = carry << 1;

    }

    return x;

}

int Add(int x, int y)

{

    if (y == 0)

        return x;

    else

        return Add( x ^ y, (x & y) << 1);

}

int subtract(int num1, int num2) {
    num2 = add(~num2, 1);
    return add(num1, num2);
}

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;
}

int multiply(int x, int y) {
    int result = 0;

    if(x < 0 && y < 0) {
        return multiply(negate(x), negate(y));
    }

    if(x >= 0 && y < 0) {
        return multiply(y, x);
    }

    while(y > 0) {
        if(is_even(y)) {
            x = multiply_by_two(x);
            y = divide_by_two(y);
        } else {
            result = add(result, x);
            y = add(y, -1);
        }
    }

    return result;
}

From Coding Interviews: Questions, Analysis & Solutions
int divide(int num1, int num2) {
    if (num2 == 0)
        throw new ArithmeticException("num2 is zero.");

    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;
    for (int i = 0; i < 32; i=add(i, 1)) {
        result = result << 1;
        if ((num1 >> (31 - i)) >= num2) {
            num1 = subtract(num1, num2 << (31 - i));
            result = add(result, 1);
        }
    }

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

    return result;
}

int divide(int num1, int num2) {
    if (num2 == 0)
        throw new ArithmeticException("num2 is zero.");

    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;
    for (int i = 0; i < 32; i=add(i, 1)) {
        result = result << 1;
        if ((num1 >> (31 - i)) >= num2) {
            num1 = subtract(num1, num2 << (31 - i));
            result = add(result, 1);
        }
    }

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

    return result;
}

int add(int x, int y) {
    int carry = 0;
    int result = 0;
    int i;

    for(i = 0; i < 32; ++i) {
        int a = (x >> i) & 1;
        int b = (y >> i) & 1;
        result |= ((a ^ b) ^ carry) << i;
        carry = (a & b) | (b & carry) | (carry & a);
    }

    return result;
}

int negate(int x) {
    return add(~x, 1);
}

int subtract(int x, int y) {
    return add(x, negate(y));
}

int is_even(int n) {
    return !(n & 1);
}

int divide_by_two(int n) {
    return n >> 1;
}

int multiply_by_two(int n) {
    return n << 1;
}

Read full article from c++ - How to add two numbers without using ++ or + or another arithmetic operator - Stack Overflow