Why does the behavior of the Integer constant pool change at 127?
Solution 1:
No, the constant pool for numbers doesn't work the same way as for strings. For strings, only compile-time constants are interned - whereas for the wrapper types for integer types, any boxing operation will always use the pool if it's applicable for that value. So for example:
int x = 10;
int y = x + 1;
Integer z = y; // Not a compile-time constant!
Integer constant = 11;
System.out.println(z == constant); // true; reference comparison
The JLS guarantees a small range of pooled values, but implementations can use a wider range if they wish.
Note that although it's not guaranteed, every implementation I've looked at uses Integer.valueOf
to perform boxing operations - so you can get the same effect without the language's help:
Integer x = Integer.valueOf(100);
Integer y = Integer.valueOf(100);
System.out.println(x == y); // true
From section 5.1.7 of the JLS:
If the value p being boxed is true, false, a byte, or a char in the range \u0000 to \u007f, or an int or short number between -128 and 127 (inclusive), then let r1 and r2 be the results of any two boxing conversions of p. It is always the case that r1 == r2.
Ideally, boxing a given primitive value p, would always yield an identical reference. In practice, this may not be feasible using existing implementation techniques. The rules above are a pragmatic compromise. The final clause above requires that certain common values always be boxed into indistinguishable objects. The implementation may cache these, lazily or eagerly. For other values, this formulation disallows any assumptions about the identity of the boxed values on the programmer's part. This would allow (but not require) sharing of some or all of these references.
This ensures that in most common cases, the behavior will be the desired one, without imposing an undue performance penalty, especially on small devices. Less memory-limited implementations might, for example, cache all char and short values, as well as int and long values in the range of -32K to +32K.
Solution 2:
Java maintains Integer pool from -128
to 127
Declaring Integer like below
Integer i1 = 127;
Results in to
Integer i1 = Integer.valueOf(127);
So what actually happening for first case is
Integer i1 = 127;<---Integer.valueOf(127);
Integer i2 = 127;<---Integer.valueOf(127);<---Same reference as first
From source code of Integer
for class valueOf
method
public static Integer valueOf(int i) {
if(i >= -128 && i <= IntegerCache.high)
return IntegerCache.cache[i + 128];
else
return new Integer(i);
}
So you get same reference if value is between -128
to 127
and you call valueOf
else it just returns new Integer(i)
And because reference is same your ==
operator works for integers returned by valueOf
between this range.
Solution 3:
Java caches the integer objects in the range -128 to 127
. So, when you try to assign a value in this range to a wrapper
object, the boxing
operation will invoke Integer.valueOf
method and in turn it will assign a reference to the object already in the pool.
On the other hand, if you assign a value outside this range to a wrapper
reference type, Integer.valueOf
will create a new Integer
object for that value. And hence, comparing the reference
for Integer
objects having value outside this range will give you false
So,
Integer i = 127; --> // Equivalent to `Integer.valueOf(127)`
Integer i2 = 127;
// Equivalent to `Integer.valueOf(128)`
// returns `new Integer(128)` for value outside the `Range - [-128, 127]`
Integer i3 = 128;
Integer i4 = 128;
System.out.println(i == i2); // true, reference pointing to same literal
System.out.println(i3 == i4); // false, reference pointing to different objects
But , when you create your integer instances using new
operator, a new object will be created on Heap.
So,
Integer i = new Integer(127);
Integer i2 = new Integer(127);
System.out.println(i == i2); // false
Solution 4:
In short newer versions of Java cache Integer is in the -128 to 127 range (256 values). look here
What exactly does comparing Integers with == do?