Comparing boxed Long values 127 and 128
Solution 1:
TL;DR
Java caches boxed Integer instances from -128
to 127
. Since you are using ==
to compare objects references instead of values, only cached objects will match. Either work with long
unboxed primitive values or use .equals()
to compare your Long
objects.
Long (pun intended) version
Why there is problem in comparing Long variable with value greater than 127? If the data type of above variable is primitive (long) then code work for all values.
Java caches Integer objects instances from the range -128 to 127. That said:
- If you set to N Long variables the value
127
(cached), the same object instance will be pointed by all references. (N variables, 1 instance) - If you set to N Long variables the value
128
(not cached), you will have an object instance pointed by every reference. (N variables, N instances)
That's why this:
Long val1 = 127L;
Long val2 = 127L;
System.out.println(val1 == val2);
Long val3 = 128L;
Long val4 = 128L;
System.out.println(val3 == val4);
Outputs this:
true
false
For the 127L value, since both references (val1 and val2) point to the same object instance in memory (cached), it returns true
.
On the other hand, for the 128 value, since there is no instance for it cached in memory, a new one is created for any new assignments for boxed values, resulting in two different instances (pointed by val3 and val4) and returning false
on the comparison between them.
That happens solely because you are comparing two Long
object references, not long
primitive values, with the ==
operator. If it wasn't for this Cache mechanism, these comparisons would always fail, so the real problem here is comparing boxed values with ==
operator.
Changing these variables to primitive long
types will prevent this from happening, but in case you need to keep your code using Long
objects, you can safely make these comparisons with the following approaches:
System.out.println(val3.equals(val4)); // true
System.out.println(val3.longValue() == val4.longValue()); // true
System.out.println((long)val3 == (long)val4); // true
(Proper null checking is necessary, even for castings)
IMO, it's always a good idea to stick with .equals() methods when dealing with Object comparisons.
Reference links:
- https://today.java.net/pub/a/today/2005/03/24/autoboxing.html
- https://blogs.oracle.com/darcy/entry/boxing_and_caches_integer_valueof
- http://java.dzone.com/articles/surprising-results-autoboxing
Solution 2:
Java caches the primitive values from -128 to 127. When we compare two Long objects java internally type cast it to primitive value and compare it. But above 127 the Long object will not get type caste. Java caches the output by .valueOf() method.
This caching works for Byte, Short, Long from -128 to 127. For Integer caching works From -128 to java.lang.Integer.IntegerCache.high or 127, whichever is bigger.(We can set top level value upto which Integer values should get cached by using java.lang.Integer.IntegerCache.high).
For example:
If we set java.lang.Integer.IntegerCache.high=500;
then values from -128 to 500 will get cached and
Integer a=498;
Integer b=499;
System.out.println(a==b)
Output will be "true".
Float and Double objects never gets cached.
Character will get cache from 0 to 127
You are comparing two objects. so == operator will check equality of object references. There are following ways to do it.
1) type cast both objects into primitive values and compare
(long)val3 == (long)val4
2) read value of object and compare
val3.longValue() == val4.longValue()
3) Use equals() method on object comparison.
val3.equals(val4);
Solution 3:
num1
and num2
are Long objects. You should be using equals()
to compare them. ==
comparison might work sometimes because of the way JVM boxes primitives, but don't depend on it.
if (num1.equals(num1))
{
//code
}
Solution 4:
Comparing non-primitives (aka Objects) in Java with ==
compares their reference instead of their values. Long
is a class and thus Long
values are Objects.
The problem is that the Java Developers wanted people to use Long
like they used long
to provide compatibility, which led to the concept of autoboxing, which is essentially the feature, that long
-values will be changed to Long
-Objects and vice versa as needed. The behaviour of autoboxing is not exactly predictable all the time though, as it is not completely specified.
So to be safe and to have predictable results always use .equals()
to compare objects and do not rely on autoboxing in this case:
Long num1 = 127, num2 = 127;
if(num1.equals(num2)) { iWillBeExecutedAlways(); }