找出数组中的两个单身狗数/C/C++

//找出单身狗
#include 

int main() {
	int arr[10] = { 1,2,3,4,5,1,2,3,4,7};
	int len = sizeof(arr) / sizeof(int);
	int i;
	int result = arr[0];
	for (i = 1; i < len; i++) {
		result = result ^ arr[i];
	}//result为非0数，且值为1的二进制位两个单生狗数不同
	int ret;
	while (result) {
		ret = result;
		result = result & (result - 1);
	}//使result二进制只有一个1，赋给ret
	int a = 0;
	int b = 0;
	for (i = 0; i < len; i++) {
		if ((arr[i] & ret) == 0) {
			a ^= arr[i];
		}
		else {
			b ^= arr[i];
		}
	}//分成两类，各自异或
	printf("单身狗分别是%d , %d\n", a, b);
	return 0;
}