StackOverflow热帖:Java整数相加溢出怎么办?

做者 | Aaron_涛
ide

来源 | blog.csdn.net/qq_33330687/article/details/81626157函数

# 问题

在以前刷题的时候碰见一个问题,须要解决int相加后怎么判断是否溢出,若是溢出就返回Integer.MAX_VALUEspa

# 解决方案

JDK8已经帮咱们实现了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~.net

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是不同的code

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;
}

如何使用?

直接调用是最方便的,可是为了追求速度,应该修改一下,理解判断思路,由于异常是十分耗时的操做,无脑异常有可能超时blog

# 写这个的目的

总结一下,也方便告诉他人Java帮咱们写好了函数。ip

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