摘要:
移位操作:x > i)) + "\n\r"); } }输出:i = 80000000 i = c0000000 i = e0000000因为:0x80000000 二进制是11100000000 ,符号位为1就在右边补1,0则补0若:System.out.println(Integer.toHexString(0x80000000 >> 32));输出:ffffffffSystem.out.println(Integer.toHexString(0x80000000 >> 32));输出为:80000000总结:0x80000000在右移31位 阅读全文