Difference between double and Double in comparison
double, java
Solution
`c` and `d` are technically two different objects and `==` operator compares only references.
c.equals(d)
is better as it compares values, not references. But still not ideal. Comparing floating-point values directly should always take some error (epsilon) into account (`Math.abs(c - d) < epsilon`).
Note that:
Integer c = 1;
Integer d = 1;
here comparison would yield `true`, but that's more complicated (`Integer` internal caching, described in JavaDoc of `Integer.valueOf()`):
This method will always cache values in the range -128 to 127, inclusive, and may cache other values outside of this range.
Why `valueOf()`? Because this method is implicitly used to implement autoboxing:
Integer c = Integer.valueOf(1);
Integer d = Integer.valueOf(1);
See also
- Weird Integer boxing in Java
- How to properly compare two Integers in Java?
Problem
I know that `Double` is a a wrapper class, and it wraps `double` number. Today, I have seen another main difference : ``` double a = 1.0; double b = 1.0; Double c = 1.0; Double d = 1.0; System.out.println(a == b); // true System.out.println(c == d); // false ``` So strange with me !!! So, if we use `Double`, each time, we must do something like this : ``` private static final double delta = 0.0001; System.out.println(Math.abs(c-d) < delta); ``` I cannot explain why Double make directly comparison wrong. Please explain for me.