问题描述
如题，我自己测试的时候，如果两个int型大数相乘，结果为负，求检测溢出的方法，并掷出异常的代码
解决方案
/** * Multiply two integers, checking for overflow. * * @param x a factor * @param y a factor * @return the product <code>x*y</code> * @throws ArithmeticException if the result can not be represented as an * int * @since 1.1 */ public static int mulAndCheck(int x, int y) { long m = ((long)x) * ((long)y); if (m < Integer.MIN_VALUE || m > Integer.MAX_VALUE) { throw new ArithmeticException("overflow: mul"); } return (int)m; } /** * Multiply two long integers, checking for overflow. * * @param a first value * @param b second value * @return the product <code>a * b</code> * @throws ArithmeticException if the result can not be represented as an * long * @since 1.2 */ public static long mulAndCheck(long a, long b) { long ret; String msg = "overflow: multiply"; if (a > b) { // use symmetry to reduce boundary cases ret = mulAndCheck(b, a); } else { if (a < 0) { if (b < 0) { // check for positive overflow with negative a, negative b if (a >= Long.MAX_VALUE / b) { ret = a * b; } else { throw new ArithmeticException(msg); } } else if (b > 0) { // check for negative overflow with negative a, positive b if (Long.MIN_VALUE / b <= a) { ret = a * b; } else { throw new ArithmeticException(msg); } } else { // assert b == 0 ret = 0; } } else if (a > 0) { // assert a > 0 // assert b > 0 // check for positive overflow with positive a, positive b if (a <= Long.MAX_VALUE / b) { ret = a * b; } else { throw new ArithmeticException(msg); } } else { // assert a == 0 ret = 0; } } return ret; }以上是commons-math-2.2的源码http://commons.apache.org/math/download_math.cgi
解决方案二：
int c = a*b;if(c/b!=a){ System.out.println("溢出");} 这是一个大概的思路，楼主可以尝试一下