StackOverflow热帖:Java整数相加溢出怎么办?Java8一步搞定~

2020-11-02 00:00:00 专区 订阅 付费 异常 溢出

作者:Aaron_涛

来源:blog.csdn.net/qq_33330687/article/details/81626157

问题


在之前刷题的时候遇见一个问题,需要解决int相加后怎么判断是否溢出,如果溢出就返回Integer.MAX_VALUE


解决方案


JDK8已经帮我们实现了Math下,不得不说这个方法是在StackOverflow找到了的,确实比国内一些论坛好多了~


加法

public static int addExact(int x, int y) { int r = x + y; // HD 2-12 Overflow iff both arguments have the opposite sign of the result if (((x ^ r) & (y ^ r)) < ) { throw new ArithmeticException("integer overflow"); } return r;}

减法

public static int subtractExact(int x, int y) { int r = x - y; // HD 2-12 Overflow iff the arguments have different signs and // the sign of the result is different than the sign of x if (((x ^ y) & (x ^ r)) < ) { throw new ArithmeticException("integer overflow"); } return r;}

乘法

public static int multiplyExact(int x, int y) { long r = (long)x * (long)y; if ((int)r != r) { throw new ArithmeticException("integer overflow"); } return (int)r;}

注意 long和int是不一样的
public static long multiplyExact(long x, long y) { long r = x * y; long ax = Math.abs(x); long ay = Math.abs(y); if (((ax | ay) >>> 31 != )) { // Some bits greater than 2^31 that might cause overflow // Check the result using the divide operator // and check for the special case of Long.MIN_VALUE * -1 if (((y != ) && (r / y != x)) || (x == Long.MIN_VALUE && y == -1)) { throw new ArithmeticException("long overflow"); } } return r;}

如何使用?


直接调用是方便的,但是为了追求速度,应该修改一下,理解判断思路,因为异常是十分耗时的操作,无脑异常有可能超时


写这个的目的


总结一下,也方便告诉他人Java帮我们写好了函数。
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