Java整数相加溢出怎么办？Java 8 还是厉害！

趣学程序-shaofeer

修改于 2020-12-03 13:49:53

5440

修改于 2020-12-03 13:49:53

文章被收录于专栏：upuptop的专栏upuptop的专栏

作者：Aaron_涛

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)) < 0) {
            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)) < 0) {
            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 != 0)) {
            // 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 != 0) && (r / y != x)) ||
               (x == Long.MIN_VALUE && y == -1)) {
                throw new ArithmeticException("long overflow");
            }
        }
        return r;
}