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# 不用加减乘除做加法
 [](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#简单) [](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#剑指offer) [](https://imhuay.gitbook.io/studies/algorithms/problems/2022/01/pages/R5NyOzkn3qAZy7wCx1pS#位运算)
**问题简述**
```
求两个整数之和，要求不能使用 “+”、“-”、“*”、“/” 运算符号。
```
详细描述
```
写一个函数，求两个整数之和，要求在函数体内不得使用 “+”、“-”、“*”、“/” 四则运算符号。
示例:
输入: a = 1, b = 1
输出: 2
提示：
a, b 均可能是负数或 0
结果不会溢出 32 位整数
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/bu-yong-jia-jian-cheng-chu-zuo-jia-fa-lcof
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
```
**思路**
> [不用加减乘除做加法（位运算，清晰图解）](https://leetcode-cn.com/problems/bu-yong-jia-jian-cheng-chu-zuo-jia-fa-lcof/solution/mian-shi-ti-65-bu-yong-jia-jian-cheng-chu-zuo-ji-7/)
* 不用编程语言之间略有区别；
Java（推荐）
```java
class Solution {
public int add(int a, int b) {
while(b != 0) { // 当进位为 0 时跳出
int c = (a & b) << 1; // c = 进位
a ^= b; // a = 非进位和
b = c; // b = 进位
}
return a;
}
}
```
Python
* Python 中
```python
class Solution:
def add(self, a: int, b: int) -> int:
x = 0xffffffff
a, b = a & x, b & x # 转为补码形式
while b != 0:
a, b = (a ^ b), (a & b) << 1 & x
return a if a <= 0x7fffffff else ~(a ^ x) # 还原
```
C++
[不用加减乘除做加法](https://leetcode-cn.com/problems/bu-yong-jia-jian-cheng-chu-zuo-jia-fa-lcof/solution/dian-zan-yao-wo-zhi-dao-ni-xiang-kan-dia-ovxy/)
```cpp
class Solution {
public:
int add(int a, int b) {
while (b) {
int carry = a & b; // 计算 进位
a = a ^ b; // 计算 本位
b = (unsigned)carry << 1; // C++中负数不支持左位移
}
return a;
}
};
```
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