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
[code="java"]
/**
* Multiply two integers, checking for overflow.
*
* @param x a factor
* @param y a factor
* @return the product x*y
* @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;
}
[/code]
以上是commons-math-2.2的源码
[url]http://commons.apache.org/math/download_math.cgi[/url]
[code="java"]
int c = a*b;
if(c/b!=a){
System.out.println("溢出");
}
[/code]
这是一个大概的思路,楼主可以尝试一下