前端突然报了integer overflow错误,int类型溢出也就是数字超过了int类型，一看很懵逼，查看后台日期发现是在Math.toIntExact()方法报错
那么我们看下方法内部代码：
/**
* Returns the value of the {@code long} argument;
* throwing an exception if the value overflows an {@code int}.
*
* @param value the long value
* @return the argument as an int
* @throws ArithmeticException if the {@code argument} overflows an int
* @since 1.8
*/
public static int toIntExact(long value) {
if ((int)value != value) {
throw new ArithmeticException("integer overflow");
}
return (int)value;
}
从代码中看出，它对参数value值进行了验证，如果强转为int仍然与原值相等说明没有超过int的值范围，否则抛出异常：integer overflow
我们在看看Math的其他方法，既然我们需要加减乘除，Math类肯定有现成的也会支持long类型的
这里只截出一部分，我们要用到的，这些都是1.8才有的方法
long类型的加减乘除运算：
加法：支持 int 和long的参数
*/
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;
} /**
* Returns the sum of its arguments,
* throwing an exception if the result overflows a {@code long}.
*
* @param x the first value
* @param y the second value
* @return the result
* @throws ArithmeticException if the result overflows a long
* @since 1.8
*/
public static long addExact(long x, long y) {
long 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("long overflow");
}
减法运算：
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;
}
/**
* Returns the difference of the arguments,
* throwing an exception if the result overflows a {@code long}.
*
* @param x the first value
* @param y the second value to subtract from the first
* @return the result
* @throws ArithmeticException if the result overflows a long
* @since 1.8
*/
public static long subtractExact(long x, long y) {
long 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("long overflow");
}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;
}
/**
* Returns the product of the arguments,
* throwing an exception if the result overflows a {@code long}.
*
* @param x the first value
* @param y the second value
* @return the result
* @throws ArithmeticException if the result overflows a long
* @since 1.8
*/
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;
}除法：除法时乡下取整的
public static int floorDiv(int x, int y) {
int r = x / y;
// if the signs are different and modulo not zero, round down
if ((x ^ y) < 0 && (r * y != x)) {
r--;
}
return r;
}
public static long floorDiv(long x, long y) {
long r = x / y;
// if the signs are different and modulo not zero, round down
if ((x ^ y) < 0 && (r * y != x)) {
r--;
}
return r;
}