47-Bitwise implementation Subtraction arithmetic

Implement two-digit arithmetic without using the +-*/arithmetic character.

1. Add

The addition of two numbers is achieved with bits. The addition of 2 numbers and the overflow problem are not considered.
such as 9+15=24
1001 + 1111, due to binary 0+0=0,1+0=1, 0+1=1, 1+1=0, can be found to be XOR operation, and produce carry, then only 1, 1 add, that is, with the operation.

int add(int nums1, int nums2) { if (nums1 == 0 || nums2 == 0) return (nums1 == 0) ? nums2 : nums1; int sum = 0, carry = 0; do { sum = nums1 ^ nums2; carry = (nums1 & nums2) << 1; // 进位左移一位 nums1 = sum; nums2 = carry; } while (nums2); return sum;}

2. Reduce

A-B = A + (-a), which requires only a negative operation, that is, complement: bitwise negation + 1

int negtive(int num) {    return1);}intsub(int nums1, int nums2) {    return add(nums1, negtive(nums2));}

3. Multiply

such as multiplication in decimal, bitwise multiply.
To find the product of 2 positive integers

int multi(int nums1, int nums2) { if (nums1 == 0 || nums2 == 0) return 0; int result = 0; while (nums2) { if (nums2 & 0x01) { // 乘数的最后1位 result = add(result, nums1); } nums1 <<= 1; // 被乘数左移1位 nums2 >>= 1; // 取乘数的下一位 } return result;}

4. In addition to

A quotient that divides two positive integers.
Multiply the inverse.
Division is the inverse of the multiplication process, X/y process: × minus (if x is reduced) y^ (2^31), y^ (2^30),... y^8,y^4,y^2,y^1. Subtract the corresponding number of y in the result plus the corresponding quantity.

intDivision (intNUMS1,intNUMS2) {if(Nums1 = =0|| NUMS2 = =0)return 0;Const intBITS =sizeof(NUMS1) *8;intresult =0; for(inti = bits-1; I >=0; i--) {//The reason for not using nums2 << I compared with nums1 is to prevent overflow        if((nums1 >> i) > Nums2) {result = Add (result,1<< i);        Nums1 = Sub (nums1, nums2 << i); }    }returnResult;}

Reference: http://blog.csdn.net/hackbuteer1/article/details/7390093

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47-Bitwise implementation Subtraction arithmetic