Java double.MAX_VALUE?

Double.MAX_VALUE is the maximum value a double can represent (somewhere around 1.7*10^308).

This should end in some calculation problems, if you try to subtract the maximum possible value of a data type.

Even though when you are dealing with money you should never use floating point values especially while rounding this can cause problems (you will either have to much or less money in your system then).

Resurrecting the dead here, but just in case someone stumbles against this like myself. I know where to get the maximum value of a double, the (more) interesting part was to how did they get to that number.

double has 64 bits. The first one is reserved for the sign.

Next 11 represent the exponent (that is 1023 biased). It's just another way to represent the positive/negative values. If there are 11 bits then the max value is 1023.

Then there are 52 bits that hold the mantissa.

This is easily computed like this for example:

public static void main(String[] args) {

    String test = Strings.repeat("1", 52);

    double first = 0.5;
    double result = 0.0;
    for (char c : test.toCharArray()) {
        result += first;
        first = first / 2;
    }

    System.out.println(result); // close approximation of 1
    System.out.println(Math.pow(2, 1023) * (1 + result));
    System.out.println(Double.MAX_VALUE);

} 

You can also prove this in reverse order :

    String max = "0" + Long.toBinaryString(Double.doubleToLongBits(Double.MAX_VALUE));

    String sign = max.substring(0, 1);
    String exponent = max.substring(1, 12); // 11111111110
    String mantissa = max.substring(12, 64);

    System.out.println(sign); // 0 - positive
    System.out.println(exponent); // 2046 - 1023 = 1023
    System.out.println(mantissa); // 0.99999...8