Compare two primitive long variables in java

in java,

use == to compare primitives and x.equals(y) to compare objects.

I'm not sure what your question is-- it looks like you tried both, and did neither work?

Note, you are using Long instead of long, so you want to use .equals()

(long is a primitive, Long is an object)

In java:

the == operator tells you if the two operands are the same object (instance).
the .equals() method on Long tells you if they are equal in value.

But you shouldn't do either. The correct way to do it is this:

assertEquals(id1, id2);

With assertEquals(), if the assertion fails, the error message will tell you what the two values were, eg expected 2, but was 5 etc

Try doing the following:

assertTrue(id1.longValue() == id2.longValue())

To compare two primitive long you can simply use ==

Example:

long x = 1L;
long y = 1L;

if (x == y) {
 System.out.println("value of x and y are same");
}

To compare two Long objects you can use Long.compare(long x, long y). This method was added in java 1.7. Below is the method implementation:

public static int compare(long x, long y) {
        return (x < y) ? -1 : ((x == y) ? 0 : 1);
}

Example:

Long x = new Long(1);
Long y = new Long(1);
if (Long.compare(x,y) == 0) {
  System.out.println(values of x and y are same);
}

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