This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Wrapper class and == operator
It seems as if object equality operator for wrapper classes produces different results depending on whether the wrapped value is in byte range or not. Here is a code snippet to demonstrate this behavior:
System.out.println("smaller than byte");
Integer i1 = 1;
Integer i2 = 1;
if (i1 == i2) System.out.println("same");
if (i1 != i2) System.out.println("not same");
System.out.println("larger than byte");
Integer i3 = 128;
Integer i4 = 128;
if (i3 == i4) System.out.println("same");
if (i3 != i4) System.out.println("not same");
produces the following output:
smaller than byte
same
larger than byte
not same
Note: I got this output on HotSpot (build 1.6.0_24-b07) on linux. Same happens for Long and probably Short (haven't tested it though).
Note: Same output on other HotSpot builds under linux
Can anyone explain it?
Small edit, just to make it slightly more interesting:
Adding
if (i3 <= i4 && i3 >= i4) System.out.println("same after all...");
in the end, prints "same after all...".
That's correct. The JVM will "cache" and reuse Integer instances when autoboxing small values.
See Java Language Specification Section 5.1.7 Boxing Conversion:
If the value p being boxed is true, false, a byte, a char in the range \u0000 to \u007f, or an int or short number between -128 and 127, then let r1 and r2 be the results of any two boxing conversions of p. It is always the case that r1 == r2.
When comparing Integers using <, >, <= and >= the values are un-boxed as opposed to != and ==.
Integers between -127 and 127 are 'cached', so they return same reference, which means i1 and i2 point to same object.
Related
This question already has answers here:
Why is 128==128 false but 127==127 is true when comparing Integer wrappers in Java?
(8 answers)
Closed 8 years ago.
Why Integer == operator does not work for 128 and after Integer values? Can someone explain this situation?
This is my Java environment:
java version "1.6.0_37"
Java(TM) SE Runtime Environment (build 1.6.0_37-b06)
Java HotSpot(TM) 64-Bit Server VM (build 20.12-b01, mixed mode)
Sample Code:
Integer a;
Integer b;
a = 129;
b = 129;
for (int i = 0; i < 200; i++) {
a = i;
b = i;
if (a != b) {
System.out.println("Value:" + i + " - Different values");
} else {
System.out.println("Value:" + i + " - Same values");
}
}
Some part of console output:
Value:124 - Same values
Value:125 - Same values
Value:126 - Same values
Value:127 - Same values
Value:128 - Different values
Value:129 - Different values
Value:130 - Different values
Value:131 - Different values
Value:132 - Different values
Check out the source code of Integer . You can see the caching of values there.
The caching happens only if you use Integer.valueOf(int), not if you use new Integer(int). The autoboxing used by you uses Integer.valueOf.
According to the JLS, you can always count on the fact that for values between -128 and 127, you get the identical Integer objects after autoboxing, and on some implementations you might get identical objects even for higher values.
Actually in Java 7 (and I think in newer versions of Java 6), the implementation of the IntegerCache class has changed, and the upper bound is no longer hardcoded, but it is configurable via the property "java.lang.Integer.IntegerCache.high", so if you run your program with the VM parameter -Djava.lang.Integer.IntegerCache.high=1000, you get "Same values" for all values.
But the JLS still guarantees it only until 127:
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 characters and shorts, as well as integers and longs in the range of -32K - +32K.
Integer is a wrapper class for int.
Integer != Integer compares the actual object reference, where int != int will compare the values.
As already stated, values -128 to 127 are cached, so the same objects are returned for those.
If outside that range, separate objects will be created so the reference will be different.
To fix it:
Make the types int or
Cast the types to int or
Use .equals()
According to Java Language Specifications:
If the value p being boxed is true, false, a byte, a char in the range
\u0000 to \u007f, or an int or short number between -128 and 127, then
let r1 and r2 be the results of any two boxing conversions of p. It is
always the case that r1 == r2.
JLS Boxing Conversions
Refer to this article for more information on int caching
The Integer object has an internal cache mechanism:
private static class IntegerCache {
static final int high;
static final Integer cache[];
static {
final int low = -128;
// high value may be configured by property
int h = 127;
if (integerCacheHighPropValue != null) {
// Use Long.decode here to avoid invoking methods that
// require Integer's autoboxing cache to be initialized
int i = Long.decode(integerCacheHighPropValue).intValue();
i = Math.max(i, 127);
// Maximum array size is Integer.MAX_VALUE
h = Math.min(i, Integer.MAX_VALUE - -low);
}
high = h;
cache = new Integer[(high - low) + 1];
int j = low;
for(int k = 0; k < cache.length; k++)
cache[k] = new Integer(j++);
}
private IntegerCache() {}
}
Also see valueOf method:
public static Integer valueOf(int i) {
if(i >= -128 && i <= IntegerCache.high)
return IntegerCache.cache[i + 128];
else
return new Integer(i);
}
This is why you should use valueOf instead of new Integer. Autoboxing uses this cache.
Also see this post: https://effective-java.com/2010/01/java-performance-tuning-with-maximizing-integer-valueofint/
Using == is not a good idea, use equals to compare the values.
Use .equals() instead of ==.
Integer values are only cached for numbers between -127 and 128, because they are used most often.
if (a.equals(b)) { ... }
Depending on how you get your Integer instances, it may not work for any value:
System.out.println(new Integer(1) == new Integer(1));
prints
false
This is because the == operator applied to reference-typed operands has nothing to do with the value those operands represent.
It's because Integer class implementation logic. It has prepared objects for numbers till 128. You can checkout http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/6-b14/java/lang/Integer.java source of open-jdk for example (search for cache[]).
Basically objects shouldn't be compared using == at all, with one exception to Enums.
This question already has answers here:
Why is 128==128 false but 127==127 is true when comparing Integer wrappers in Java?
(8 answers)
Closed 8 years ago.
Why Integer == operator does not work for 128 and after Integer values? Can someone explain this situation?
This is my Java environment:
java version "1.6.0_37"
Java(TM) SE Runtime Environment (build 1.6.0_37-b06)
Java HotSpot(TM) 64-Bit Server VM (build 20.12-b01, mixed mode)
Sample Code:
Integer a;
Integer b;
a = 129;
b = 129;
for (int i = 0; i < 200; i++) {
a = i;
b = i;
if (a != b) {
System.out.println("Value:" + i + " - Different values");
} else {
System.out.println("Value:" + i + " - Same values");
}
}
Some part of console output:
Value:124 - Same values
Value:125 - Same values
Value:126 - Same values
Value:127 - Same values
Value:128 - Different values
Value:129 - Different values
Value:130 - Different values
Value:131 - Different values
Value:132 - Different values
Check out the source code of Integer . You can see the caching of values there.
The caching happens only if you use Integer.valueOf(int), not if you use new Integer(int). The autoboxing used by you uses Integer.valueOf.
According to the JLS, you can always count on the fact that for values between -128 and 127, you get the identical Integer objects after autoboxing, and on some implementations you might get identical objects even for higher values.
Actually in Java 7 (and I think in newer versions of Java 6), the implementation of the IntegerCache class has changed, and the upper bound is no longer hardcoded, but it is configurable via the property "java.lang.Integer.IntegerCache.high", so if you run your program with the VM parameter -Djava.lang.Integer.IntegerCache.high=1000, you get "Same values" for all values.
But the JLS still guarantees it only until 127:
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 characters and shorts, as well as integers and longs in the range of -32K - +32K.
Integer is a wrapper class for int.
Integer != Integer compares the actual object reference, where int != int will compare the values.
As already stated, values -128 to 127 are cached, so the same objects are returned for those.
If outside that range, separate objects will be created so the reference will be different.
To fix it:
Make the types int or
Cast the types to int or
Use .equals()
According to Java Language Specifications:
If the value p being boxed is true, false, a byte, a char in the range
\u0000 to \u007f, or an int or short number between -128 and 127, then
let r1 and r2 be the results of any two boxing conversions of p. It is
always the case that r1 == r2.
JLS Boxing Conversions
Refer to this article for more information on int caching
The Integer object has an internal cache mechanism:
private static class IntegerCache {
static final int high;
static final Integer cache[];
static {
final int low = -128;
// high value may be configured by property
int h = 127;
if (integerCacheHighPropValue != null) {
// Use Long.decode here to avoid invoking methods that
// require Integer's autoboxing cache to be initialized
int i = Long.decode(integerCacheHighPropValue).intValue();
i = Math.max(i, 127);
// Maximum array size is Integer.MAX_VALUE
h = Math.min(i, Integer.MAX_VALUE - -low);
}
high = h;
cache = new Integer[(high - low) + 1];
int j = low;
for(int k = 0; k < cache.length; k++)
cache[k] = new Integer(j++);
}
private IntegerCache() {}
}
Also see valueOf method:
public static Integer valueOf(int i) {
if(i >= -128 && i <= IntegerCache.high)
return IntegerCache.cache[i + 128];
else
return new Integer(i);
}
This is why you should use valueOf instead of new Integer. Autoboxing uses this cache.
Also see this post: https://effective-java.com/2010/01/java-performance-tuning-with-maximizing-integer-valueofint/
Using == is not a good idea, use equals to compare the values.
Use .equals() instead of ==.
Integer values are only cached for numbers between -127 and 128, because they are used most often.
if (a.equals(b)) { ... }
Depending on how you get your Integer instances, it may not work for any value:
System.out.println(new Integer(1) == new Integer(1));
prints
false
This is because the == operator applied to reference-typed operands has nothing to do with the value those operands represent.
It's because Integer class implementation logic. It has prepared objects for numbers till 128. You can checkout http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/6-b14/java/lang/Integer.java source of open-jdk for example (search for cache[]).
Basically objects shouldn't be compared using == at all, with one exception to Enums.
This question already has answers here:
Weird Integer boxing in Java
(12 answers)
Closed 7 years ago.
I know int range is -2147483648 to +2147483647 but here I'm getting output as true and false. Why? Actually i1 and i2 point to the same object, so output is true. I can understood but i3 and i4 also pointing to same object but I got output as false. Why?
public class MainClass {
public static void main(String[] args) {
Integer i1 = 127;
Integer i2 = 127;
System.out.println(i1 == i2);
Integer i3 = 128;
Integer i4 = 128;
System.out.println(i3 == i4);
}
}
the output is
true
false
why output should be like this?
Because you are using Integer object. For Integer object, values in between -128 to 127 are pooled
The problem is the difference between == and equals. == just tests to see if two Integer variables point to the same object, which could be either true or false depending on the implementation of your JVM. equals actually tests to see if they hold the same value. So in this case, you'll want to use equals:
// ...
System.out.println(i1.equals(i2));
// ...
System.out.println(i3.equals(i4));
This question already has answers here:
Integer wrapper objects share the same instances only within the value 127? [duplicate]
(5 answers)
Closed 10 years ago.
Someone postulated in some forum thread that many people and even experienced Java Developers wouldn't understand the following peace of Java Code.
Integer i1 = 127;
Integer i2 = 127;
System.out.println(i1++ == i2++);
System.out.println(i1 == i2);
As a person with some interest in Java I gave it my thoughts and came to the following result.
System.out.println(i1++ == i2++);
// True, since we first check for equality and increment both variables afterwards.
System.out.println(i1 == i2);
// True again, since both variables are already incremented and have the value 128
Eclipse tells me otherwise. The first line is true and the second is false.
I really would appreciate an explanation.
Second question. Is this Java specific or does this example hold for example for the C based languages, too?
Integer i1 = 127;
Integer i2 = 127;
System.out.println(i1++ == i2++);
// here i1 and i2 are still 127 as you expected thus true
System.out.println(i1 == i2);
// here i1 and i2 are 128 which are equal but not cached
(caching range is -128 to 127),
In case 2 if you use equals() it'd return true as == operator for integers only works for cached values. as 128 is out of cache range the values above 128 will not be cached, thus youhave to use equals() method to check if two integer instances above 127 are true
TEST:
Integer i1 = 126;
Integer i2 = 126;
System.out.println(i1++ == i2++);// true
System.out.println(i1 == i2); //true
Integer i1 = 126;
Integer i2 = 126;
System.out.println(i1++ == i2++);// true
System.out.println(i1.equals(i2)); //true
Integer i1 = 128;
Integer i2 = 128;
System.out.println(i1++ == i2++);// false
System.out.println(i1==i2); //false
Integer i1 = 128;
Integer i2 = 128;
System.out.println(i1++.equals(i2++));// true
System.out.println(i1.equals(i2)); //true
As explained this is due to Integer caching. For fun, you can run the program with the following JVM option:
-XX:AutoBoxCacheMax=128
and it will print true twice (option available on hostpot 7 - not necessarily on other JVMs).
Note that:
this is JVM specific
the modified behaviour is compliant with the JLS which says that all values between -128 and +127 must be cached but also says that other values may be cached.
Bottom line: the second print statement is undefined in Java and can print true or false depending on the JVM implementation and/or JVM options used.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Inconsistent behavior on java's ==
Integer wrapper objects share the same instances only within the value 127?
I have found the following == behaviour for Integer objects and I fail to understand it. (I am well aware that one should use equals for such comparisons, but I am studying for OCPJP...)
On short, == works as expected for 1000, but not for 10.
The former fragment of code is:
Integer i1 = 1000;
Integer i2 = 1000;
if(i1 != i2) System.out.println("different objects");
if(i1.equals(i2)) System.out.println("meaningfully equal");
and it behaves as one would expect:
different objects
meaningfully equal
The latter though:
Integer i3 = 10;
Integer i4 = 10;
if(i3 == i4) System.out.println("same object");
if(i3.equals(i4)) System.out.println("meaningfully equal");
has the following output:
same object
meaningfully equal
Can someone please explain why this is happening?
Since Java 5, wrapper class caching was introduced. The following is an examination of the cache created by an inner class, IntegerCache, located in the Integer cache. For example, the following code will create a cache:
Integer myNumber = 10
or
Integer myNumber = Integer.valueOf(10);
256 Integer objects are created in the range of -128 to 127 which are all stored in an Integer array. This caching functionality can be seen by looking at the inner class, IntegerCache, which is found in Integer:
So when creating an object using Integer.valueOf or directly assigning a value to an Integer within the range of -128 to 127 the same object will be returned. Therefore, consider the following example:
Integer i = 100;
Integer p = 100;
if (i == p)
System.out.println("i and p are the same.");
if (i != p)
System.out.println("i and p are different.");
if(i.equals(p))
System.out.println("i and p contain the same value.");
The output is:
i and p are the same.
i and p contain the same value.
It is important to note that object i and p only equate to true because they are the same object, the comparison is not based on the value, it is based on object equality. If Integer i and p are outside the range of -128 or 127 the cache is not used, therefore new objects are created. When doing a comparison for value always use the “.equals” method. It is also important to note that instantiating an Integer does not create this caching.
Remember that “==” is always used for object equality, it has not been overloaded for comparing unboxed values
See Integer wrapper objects share the same instances only within the value 127?. The Integer class keeps shared static instances around for common values.