这本是careerup的一道题，看到了以后自己做了一下，主要的难点就是加法里面的进位。直接上代码：
1 int add(int a, int b) 2 { 3 const int bitsize = sizeof(int) * 8; 4 int digit[bitsize] = {0}; 5 int first = a, second = b; 6 int carry =0; 7 int result = 0; 8 for (int i = 0; i < bitsize; i++) 9 { 10 11 digit[i] = (first & 1) ^ (second & 1) ^ carry; 12 13 if (((first & 1) == 1 && (second & 1) == 1) || ((first & 1) == 1 && carry == 1) || (carry == 1 && (second & 1) == 1)) 14 carry = 1; 15 else 16 carry = 0; 17 18 first = first >> 1; 19 second = second >> 1; 20 21 result = result | (digit[i] << i); 22 } 23 return result; 24 25 }
后来网上发现了一个更加简便的方法：
1 int add_no_arithm(int a, int b) 2 { 3 if (b == 0) return a; 4 int sum = a ^ b; // add without carrying 5 int carry = (a & b) << 1; // carry, but don’t add 6 return add_no_arithm(sum, carry); // recurse 7 }
真是巧妙啊，carry就是需要进位的地方。