今天做leetcode题目时，发现int mid = 536848900，但是mid*mid=484528144，我说为啥程序运行一直出错呢！！

     int mid = 536848900;
        System.out.println(mid * mid);//484528144

        long mid2 = 536848900;//long * long
        System.out.println(mid2 * mid2);//288206741431210000

        System.out.println(Integer.MAX_VALUE);//2147483647
        System.out.println(Math.pow(2, 31) - 1);//2.147483647E9

        int a = Integer.MAX_VALUE;
//        a += 1;//-2147483648
        a += 100;//-2147483549

        long MonthNanoSeconds1 = 30 * 24 * 3600 * 1000 * 1000;//右边的结果是int,已溢出

        //int * long,与最后一个long型的1000相乘之前就已经溢出，所以结果也不对
        long MonthNanoSeconds2 = 30 * 24 * 3600 * 1000 * 1000L;

        //这是正确的写法
        long MonthNanoSeconds3 = 30L * 24 * 3600 * 1000 * 1000;//long * int

        System.out.println(MonthNanoSeconds1);//2134720512
        System.out.println(MonthNanoSeconds2);//-1702967296000
        System.out.println(MonthNanoSeconds3);//2592000000000
        System.out.println(30 * 24 * 3600 * 1000);//-1702967296
        System.out.println(30 * 24 * 3600 * 1000 * 1000);//2134720512

在Java Language Specifictionz中所述(JSL 15.7.1)

If an integer multiplication overflows, then the result is the low-order bits of the mathematical product as represented in some sufficiently large two’s-complement format. As a result, if overflow occurs, then the sign of the result may not be the same as the sign of the mathematical product of the two operand values.

也就是说int型整数相乘，结果只会保留低32位，高位会抛弃掉。

加减乘除的运算符优先级都是从左向右。

参考：

《java int溢出总结》 https://njucz.github.io/2017/08/16/java-int%E6%BA%A2%E5%87%BA%E6%80%BB%E7%BB%93/ 
《java中short、int、long、float、double取值范围》 https://blog.csdn.net/qfikh/article/details/52832087 
《[Java]运算符优先级参考图表》 https://blog.csdn.net/xiaoli_feng/article/details/4567184