一个问题~~~有关java基础~~overflow，等待~~

[cxhz_cn]

public class overflow {

public static void main(String arg[]){
int big = 0x7fffffff;
int bigger =0;
for (int i=0;i<10;i++){
bigger= big * i;
System.out.println("big"+big);
System.out.println("bigger"+bigger);
}
}
}


结果如下：
big2147483647
bigger0
big2147483647
bigger2147483647
big2147483647
bigger-2
big2147483647
bigger2147483645
big2147483647
bigger-4
big2147483647
bigger2147483643
big2147483647
bigger-6
big2147483647
bigger2147483641
big2147483647
bigger-8
big2147483647
bigger2147483639

为啥会有这样的答案~~
在i是奇数倍的时候~~~

[Luke_cn]

对于int类型来说当超出其范围时，不会抛出一个异常。
正数的最高位是0，当超出后，0变为1，此时就会变成负数。

[heifei]

0x7fffffff + 1 = 0x80000000
也就是首位为1区域为零,80000000换位十进制2147483648
0x80000000为-2147483648
类推(0x7fffffff*2)=0x7fffffff + 0x7fffffff=-2

[cxhz_cn]

给分了