Eric Sun

摘要： 这个问题不难，主要有两种方式实现。1）：用“加减”运算实现 int a = 100; int b = 200; a = a + b; //300 b = a - b; //100 a = a - b; //2002）：用“异或”运算实现 int a = 1; int b = 10; a = a ^ b; //0001 ^ 1010 = 1011 (11) b = a ^ b; //1011... 阅读全文