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

2020-11-02 09:36:04

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;}