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：
1 /** 2 * Returns the value of the {@code long} argument; 3 * throwing an exception if the value overflows an {@code int}. 4 * 5 * @param value the long value 6 * @return the argument as an int 7 * @throws ArithmeticException if the {@code argument} overflows an int 8 * @since 1.8 9 */ 10 public static int toIntExact(long value) { 11 if ((int)value != value) { 12 throw new ArithmeticException("integer overflow"); 13 } 14 return (int)value; 15 }
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
1 */ 2 public static int addExact(int x, int y) { 3 int r = x + y; 4 // HD 2-12 Overflow iff both arguments have the opposite sign of the result 5 if (((x ^ r) & (y ^ r)) < 0) { 6 throw new ArithmeticException("integer overflow"); 7 } 8 return r; 9 } 10 11 /** 12 * Returns the sum of its arguments, 13 * throwing an exception if the result overflows a {@code long}. 14 * 15 * @param x the first value 16 * @param y the second value 17 * @return the result 18 * @throws ArithmeticException if the result overflows a long 19 * @since 1.8 20 */ 21 public static long addExact(long x, long y) { 22 long r = x + y; 23 // HD 2-12 Overflow iff both arguments have the opposite sign of the result 24 if (((x ^ r) & (y ^ r)) < 0) { 25 throw new ArithmeticException("long overflow"); 26 }
减法运算：
1 public static int subtractExact(int x, int y) { 2 int r = x - y; 3 // HD 2-12 Overflow iff the arguments have different signs and 4 // the sign of the result is different than the sign of x 5 if (((x ^ y) & (x ^ r)) < 0) { 6 throw new ArithmeticException("integer overflow"); 7 } 8 return r; 9 } 10 11 /** 12 * Returns the difference of the arguments, 13 * throwing an exception if the result overflows a {@code long}. 14 * 15 * @param x the first value 16 * @param y the second value to subtract from the first 17 * @return the result 18 * @throws ArithmeticException if the result overflows a long 19 * @since 1.8 20 */ 21 public static long subtractExact(long x, long y) { 22 long r = x - y; 23 // HD 2-12 Overflow iff the arguments have different signs and 24 // the sign of the result is different than the sign of x 25 if (((x ^ y) & (x ^ r)) < 0) { 26 throw new ArithmeticException("long overflow"); 27 }
乘法：
1 public static int multiplyExact(int x, int y) { 2 long r = (long)x * (long)y; 3 if ((int)r != r) { 4 throw new ArithmeticException("integer overflow"); 5 } 6 return (int)r; 7 } 8 9 /** 10 * Returns the product of the arguments, 11 * throwing an exception if the result overflows a {@code long}. 12 * 13 * @param x the first value 14 * @param y the second value 15 * @return the result 16 * @throws ArithmeticException if the result overflows a long 17 * @since 1.8 18 */ 19 public static long multiplyExact(long x, long y) { 20 long r = x * y; 21 long ax = Math.abs(x); 22 long ay = Math.abs(y); 23 if (((ax | ay) >>> 31 != 0)) { 24 // Some bits greater than 2^31 that might cause overflow 25 // Check the result using the divide operator 26 // and check for the special case of Long.MIN_VALUE * -1 27 if (((y != 0) && (r / y != x)) || 28 (x == Long.MIN_VALUE && y == -1)) { 29 throw new ArithmeticException("long overflow"); 30 } 31 } 32 return r; 33 }
除法：除法时乡下取整的
1 public static int floorDiv(int x, int y) { 2 int r = x / y; 3 // if the signs are different and modulo not zero, round down 4 if ((x ^ y) < 0 && (r * y != x)) { 5 r--; 6 } 7 return r; 8 } 9 10 public static long floorDiv(long x, long y) { 11 long r = x / y; 12 // if the signs are different and modulo not zero, round down 13 if ((x ^ y) < 0 && (r * y != x)) { 14 r--; 15 } 16 return r; 17 }
