2使用XOR对给定字符串的补充?
在本节中,我们将看到如何使用二进制字符串的XOR操作找到2的补码。2的补码实际上是1的补码+1。我们将使用XOR操作获得1的补码。
我们将遍历LSb中的字符串,并查找0。将所有1翻转为0,直到得到0。然后翻转找到的0。
我们将从LSb出发。然后忽略所有0,直到得到1。忽略第一个1,我们将使用XOR操作切换所有位。
算法
get2sComp(bin)
begin len := length of the binary string flag := false for i := len-1 down to 0, do if bin[i] is 0, and flag is not set, then ignore the next part, jump to next iteration else if flag is set, then bin[i] := flip of bin[i] end if flag := true end if done if the flag is not set, then attach 1 with bin and return else return bin end if end
示例
#include <iostream> using namespace std; string get2sComplement(string bin) { int n = bin.length(); bool flag = false; //flag is used if 1 is seen for (int i = n - 1; i >= 0; i--) { //traverse from last bit if (bin[i] == '0' && !flag) { continue; } else { if (flag) bin[i] = (bin[i] - '0') ^ 1 + '0'; //flip bit using XOR, then convert to ASCII flag = true; } } if (!flag) //if no 1 is there, just insert 1 return "1" + bin; else return bin; } int main() { string str; cout << "Enter a binary string: "; cin >> str; cout << "2's complement of " << str <<" is " << get2sComplement(str); }
输出结果
Enter a binary string: 10110110 2's complement of 10110110 is 01001010