Java 6 - Creating and detecting the first double value above Long.MAX_VALUE

问题

Below I attempt to assign to value the maximum Long value, then add to it the minimum positive Double value possible.

I then try to detect that the value is greater than the maximum Long value.

Double value = new Long(Long.MAX_VALUE).doubleValue();
value += Double.MIN_VALUE;
if (value < -(new Long(Long.MAX_VALUE).doubleValue()) || value > new Long(Long.MAX_VALUE).doubleValue()) {
  // Expecting code here to execute, but it doesn't.
}

Studying the values involved shows that value has the final value of

9.223372036854776E18
= 9223372036854776000

while Long.MAX_VALUE has the value

9223372036854775807

Comparing these shows that value is greater as expected:

9223372036854776000 (value)
9223372036854775807 (Long.MAX_VALUE)

Could someone please explain why the if statement fails to detect this?

Sincere thanks.

回答1:

Long.MAX_VALUE can't be represented exactly as a double. The closest available double value is 9223372036854775808.

Adding Double.MIN_VALUE to that number (10^-1074) does not change anything because the closest double value to 9223372036854775808 + 10^-1074 still is 9223372036854775808.

Using @PeterLawrey's code, you can see that the next available double is 9223372036854777856 (which is equal to Long.MAX_VALUE + 2048).

Note: your double is actually equal to 9223372036854775808, which gets printed as 9.223372036854776E18. You can see it with the following code:

double d = Long.MAX_VALUE;
BigDecimal bd= new BigDecimal(d);
System.out.println(bd);

回答2:

Adding a very small value to a large value is the same as adding nothing.

I suspect what you had in mind is

double d = Long.MAX_VALUE;
// get the next represented value.
double d1 = Double.longBitsToDouble(Double.doubleToLongBits(d)+1);
System.out.println(d > Long.MAX_VALUE);
System.out.println(d1 > Long.MAX_VALUE);

prints

false
true

来源：https://stackoverflow.com/questions/12355839/java-6-creating-and-detecting-the-first-double-value-above-long-max-value