【C++】位运算实现加减乘除

 #include<iostream>
 #include<assert.h>
 using namespace std;

 // 位运算实现加减乘除

 int myAdd(int num1, int num2)
 {
     if (num2 == )    return num1;
     int sum = , carry = ;
     sum = num1 ^ num2;    // 按位抑或
     carry = (num1 & num2) << ;
     return myAdd(sum, carry);
 }

 int myAddIter(int num1, int num2)
 {
     int ret = ;
     while (num2)
     {
         ret = num1 ^ num2;
         num2 = (num1 & num2) << ;
         num1 = ret;
     }
     return ret;
 }

 int myNeg(int num1)
 {
     // 正数到负数的补码表示： 取反加1
     return myAdd(~num1, );
 }

 int myMinus(int num1, int num2)
 {
     return myAdd(num1, myNeg(num2));
 }

 // (2^0)*k0 + (2^1)*k1 + (2 ^2)*k2 + ... + (2^31)*k31
 int myPosMulti(int num1, int num2)
 {
     int ret = ;
     while (num2)
     {
         if (num2 & )    // 对应当前num2的最低位是1
             ret = myAdd(ret, num1);
         num1 <<= ;    // num1 = num1 * 2
         num2 >>= ;    // num2的当前位用过了就右移
     }
     return ret;
 }

 int myMultiply(int num1, int num2)
 {
     if (num1 >=  && num2 >= )
         return myPosMulti(num1, num2);
     if (num1 <  && num2 >= )
         return myNeg(myPosMulti(myNeg(num1), num2));
     if (num1 >=  && num2 < )
         return myNeg(myPosMulti(num1, myNeg(num2)));
     if (num1 <  && num2 < )
         return myPosMulti(myNeg(num1), myNeg(num2));
 }

 // 除法就是由乘法的过程逆推，依次减掉（如果x够减的）y^(2^31),y^(2^30),...y^8,y^4,y^2,y^1。减掉相应数量的y就在结果加上相应的数量。
 int myPosDiv(int num1, int num2)
 {
     num1 = num1 >  ? num1 : myNeg(num1);
     num2 = num2 >  ? num2 : myNeg(num2);
     if (num1 < num2)
         return ;
     long long op = num2;
     int ret = ;
     int flag;
     int curBit = ;
     assert(num2 != );
     while (op <= num1)
     {
         op <<= ;
         ++curBit;
     }
     op >>= ;
     --curBit;

     while (op >= num2)
     {
         if (num1 >= op)
         {
             num1 -= op;
             ret +=  << curBit;
         }
         op >>= ;
         --curBit;
     }
     return ret;
 }

 int myPosDivByMinus(int num1, int num2)
 {
     int ret = ;
     while (num1 >= num2)
     {
         num1 = myMinus(num1, num2);
         ret = myAdd(ret, );
     }
     return ret;
 }

 int myPosComp(int num1, int num2)
 {
     bool isZero(int);
     int temp = ;
     num2 = num1 ^ num2;
     if (isZero(num2))    return ;
     while (num2 >>= )
         temp <<= ;
     return temp & num1;
 }

 int myDivide(int num1, int num2)
 {
     if ((num1 >  && num2 > ) || (num1 <  && num2 < ))
         return myPosDiv(num1, num2);
     else
         return myNeg(myPosDiv(num1,num2));
 }

 bool isNeg(int num)
 {
     return num & 0x8000;
 }

 bool isZero(int num)
 {
     return !(num & 0xFFFF);
 }

 bool isPos(int num)
 {
     return !(num & 0x8000) && (num & 0xFFFF);
 }

 //int myMod(int a, int b)
 //{
 //
 //}

 int divide(int dividend, int divisor) {
     int ret = ;
     long long curDividend, curDivisor, op;
     bool flag;
     int curBit = ; // 记录商的1在哪一位
     if (divisor == )   return INT_MAX; // 往负数方向溢出咋整
     if ((dividend <  && divisor > ) || (dividend >  && divisor < ))
         flag = ;   // 负数
     else
         flag = ;

     curDividend = abs(dividend);
     curDivisor = abs(divisor);

     if (curDividend < curDivisor)   return ;

     op = curDivisor;
     while (op <= curDividend)
     {
         ++curBit;
         op = op << ;
     }
     op = op >> ;
     --curBit;

     while (op >= curDivisor)
     {
         if (curDividend >= op)
         {
             curDividend -= op;
             ret +=  << curBit;
         }
         op = op >> ;
         --curBit;
     }

     if (flag)
         return ret;
     return -ret;
 }

 int main()
 {
     //INT_MIN -2147483648 INT_MIN 2147483647
     int a = INT_MAX;    //
     int b = -;
     cout << "myAdd: " << myAdd(a, b) << endl;
     cout << "myMinus: " << myMinus(a, b) << endl;
     cout << "myMultiply: " << myMultiply(a, b) << endl;
     cout << "myDivide: " << myDivide(a, b) << endl;
     // -1010369383, -2147483648
     cout << "Leetcode: " << divide(-, INT_MIN) << endl;
     //cout << "myMod: " << myMod(a, b) << endl;
     system("pause");
     return ;
 }