關於integer overflow錯誤

前端突然報了integer overflow錯誤,int類型溢出也就是數字超過了int類型，一看很懵逼，查看后台日期發現是在Math.toIntExact()方法報錯

那么我們看下方法內部代碼：

/**
     * 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;
    }

從代碼中看出，它對參數value值進行了驗證，如果強轉為int仍然與原值相等說明沒有超過int的值范圍，否則拋出異常：integer overflow

我們在看看Math的其他方法，既然我們需要加減乘除，Math類肯定有現成的也會支持long類型的

這里只截出一部分，我們要用到的，這些都是1.8才有的方法

 

long類型的加減乘除運算：

加法：支持 int 和long的參數

  */
    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");
        }

減法運算：

 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：除法時鄉下取整的

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;
    }