一、递归解法：

#include<iostream>
using namespace std;
int add(int x, int y)
{
    //异或实现两数相加不进位
    int a = x^y;
    //相与移位实现进位
    int b = (x&y)<<1;
    //如果b==0，没有进位了，就返回相加的结果
    if (b == 0)
    {
        return a;
    }
    //继续递归相加和进位
    return add(a, b);
}
int main()
{
    int x, y;

    while (cin >> x >> y) {
        cout << add(x, y) << endl;
    }
    system("pause");
    return 0;
}

二、循环解法：

#include <iostream>
using namespace std;

int add1(int x, int y)
{
    int sum;
    int carry;
    int bx, by;
    int base;

    base = 1;
    //carry用来判断是否需要进位
    carry = 0;
    sum = 0;
    while (base != 0) {
        bx = x & base;
        by = y & base;
        base <<= 1;
        sum |= ((bx) ^ (by) ^ carry);
        carry = ((bx & by) || (bx & carry) || (by & carry)) ? base : 0;
    }

    return sum;
}