Java operator == is used for reference comparison
then how can == be used for comparing int a =1; and int b = 1;
both values are stored in different locations then how it compares
As commented by Andy, the JLS states that the operator '==' is indeed used for reference type comparison but also for numeric type and Boolean type comparison.
int is a numeric type.
When comparing numeric types the values are compared (not the references).
However if you want to determine if the references of two integers are equal rather than the values then you can use the Integer class. This class simply wraps the primitive numeric type int.
Now consider the following code:
public class TestClass {
public static void main(String[] args)
{
Integer A = new Integer(1);
Integer B = new Integer(1);
Integer C = A;
if (A == B) System.out.println("Won't print."); // (1)
if (A.equals(B)) System.out.println("WILL Print!!!"); // (2)
if (A == C) System.out.println("WILL Print!!!"); // (3)
}
}
Because A and B are objects, the reference of A is compared to the reference of B. Even though their int values are the same, because they are independent references this statement is false.
The equals method compares the int values of each Integer object and thus is true.
Integer object C references object A. Hence this reference comparison will be true.
Related
To my understanding, the following code should print true, since both elements are equal.
From java docs Array.get() will return:
Returns the value of the indexed component in the specified array
object. The value is automatically wrapped in an object if it has a
primitive type.
However, when I run the following code it is printing
false:
public class Test1 {
static boolean equalTest(Object array1, Object array2) {
return Array.get(array1, 0).equals(Array.get(array2, 0));
}
public static void main(String[] args) {
int[] a = new int[1];
byte[] b = new byte[1];
a[0] = 3;
b[0] = 3;
System.out.println(equalTest(a, b));
}
}
My question is isn't classes implementing Number are or should be directly comparable to one another.
This has nothing to do with arrays really. Your comparison is equivalent to:
Object x = Integer.valueOf(3);
Object y = Byte.valueOf((byte) 3);
boolean equal = x.equals(y);
That's never going to return true.
Even though your original arrays are of the primitive types, Array.get returns Object, so you're getting the boxed types - and comparing values of those different types.
According to the documentation of Array.get(Object array,int index) method, the value returned is automatically wrapped in an object if it has a primitive type. So, if you add the following lines:
System.out.println(Array.get(array1, 0).getClass());
System.out.println(Array.get(array2, 0).getClass());
you will see the output is
class java.lang.Integer
class java.lang.Byte
The equals method of Integer class first of all checks if the object it is being compared to is also an instance of Integer, if not , then no further checks are required, they are not equal.
That's the reason you see the output is false as the objects being compared for equality are Integer and Byte.
The Array.get invocations return Integer and Byte object instances. These are not equal according to Integer.equals because the class type differs.
The problem is that the Array.get() method returns an object, so for int it'll return Integer, and for byte, it will return Byte. So .equals() method will return false because it first sees whether the type is the same, and then compares the values.
The array data types don't match. One is int and the other is byte. Since they're being passed to the function as Objects, they'll end up being Integer and Byte. So in order to compare them properly, you need to type-cast them to the same type. Something like:
static boolean equalTest(Object array1, Object array2) {
int int1 = (int) Array.get(array1, 0);
int int2 = (int) Array.get(array2, 0);
return int1.equals(int2);
// OR return int1 == int2;
}
If you look at the JavaDoc for Array.get you'll see:
The value is automatically wrapped in an object if it has a primitive
type.
So your bytes become Bytes and your ints become Integers.
That means the function you're calling is Integer.equals(Object)
This is implemented like so:
public boolean equals(Object obj) {
if (obj instanceof Integer) {
return value == ((Integer)obj).intValue();
}
return false;
}
You're not passing it an Integer, you're passing it a Byte so that's why it returns false.
As we know that == in case of Objects returns true if pointing to the same reference else it returns false.
So , if i have taken
Integer a = new Integer("1"); // Creating Integer Object a
Integer b = new Integer("1"); // Creating Integer Object b
and then perform a == b , then it returns true
but they both have different references.
The JVM caches integer values between -127 to 127.
That's the reason == works for Integer value between this range.
But:
Integer i1=new Integer("11");
Integer i2=new Integer("11");
System.out.println(i1==i2); //false
Integer i3=11;
Integer i4=11;
System.out.println(i3==i4); //true
Integer i5=128;
Integer i6=128;
System.out.println(i5==i6); //false
Because it is overridden. source:
public boolean equals(Object obj) {
if (obj instanceof Integer) {
return value == ((Integer)obj).intValue();
}
return false;
}
Let me more elaborate on this, #asAmit Bhati said in his answer when we write
Integer i1=new Integer("11");
Integer i2=new Integer("11");
System.out.println(i1==i2); //false
jvm creates two separate objects and comparing them with "==" results in false. but when we write the following code:
Integer i3=11;
Integer i4=11;
System.out.println(i3==i4); //true
it will be translated into this:
Integer i3=Integer.valueOf(11);
Integer i4=Integer.valueOf(11);
Implementation of valueOf method is as follow(in java 1.8):
public static Integer valueOf(int var0) {
return var0 >= -128 && var0 <= Integer.IntegerCache.high?Integer.IntegerCache.cache[var0 + 128]:new Integer(var0);
}
as you can see if the value is between -128 and Maximum cache value (which can be configured using this jvm parameter -Djava.lang.Integer.IntegerCache.high), it will retrieve the cached value and doesn't create a new instance of Integer that is why (==) returns true for certain values!
also note that the same goes for Character Wrapper class but not for Float and Double classes.
As the answer above says, it may work in many cases, nevertheless you shouldn't compare two Integer with == since it may present problems in some cases.
Check this answer for more information:
Comparing Integer values in Java, strange behavior
If you check equals method implementation in Integer class it is:
public boolean equals(Object obj) {
if (obj instanceof Integer) {
return value == ((Integer)obj).intValue();
}
return false;
}
From here you can see it uses "==" operator.
Now the reason behind it. Ultimately you have to compare value of the Integer wrapper class, which will happen automatically because of autoboxing and unboxing in Java.
And also from the method definition you can see that it is retrieving passed Object value using ((Integer)obj).intValue().
I have two values to be compared. They appear to be not equal when compared directly using the == operator, but equal after they are put into local variables. Could anyone tell me why?
The code is as follows:
(MgrBean.getByName(name1).getId() == MgrBean.getByName(name2).getId()),
//This one is false,
int a = MgrBean.getByName(name1).getId();
int b = MgrBean.getByName(name2).getId();
(a == b); //This one is true.
I believe your getId() is returning an Integer instead of int.
Integer is the wrapper class for int primitive type. If you compare object reference using ==, you are comparing if they are referring to same object, instead of comparing the value inside. Hence if you are comparing two Integer objects reference, you are not comparing its integer value, you are simply checking if that two reference is pointing to same Integer object.
However, if you use a primitive type to store that value (e.g. int a = ....getId()), there is auto-unboxing happening, and that wrapper object is converted to primitive type value, and comparing primitive type values using == is comparing the value, which do the work you expect.
Suggested further reading for you:
Primitive Type vs Reference Type in Java
Auto-boxing and unboxing in Java
The most misconception about Integer type happens everywhere including other answers here
Suppose you have two integers defined as follows:
Integer i1 = 128
Integer i2 = 128
If you then execute the test (i1 == i2), the returned result is false. The reason is that the JVM maintains a pool of Integer values (similar to the one it maintains for Strings). But the pool contains only integers from -128 to 127. Creating any Integer in that range results in Java assigning those Integers from the pool, so the equivalency test works. However, for values greater than 127 and less than -128), the pool does not come into play, so the two assignments create different objects, which then fail the equivalency test and return false.
In contast, consider the following assignments:
Integer i1 = new Integer(1); // Any number in the Integer pool range
Integer i2 = new Integer(1); // Any number in the Integer pool range
Because these assignments are made using the 'new' keyword, they are new instances created, and not picked up from the pool. Hence, testing (i1== i2) on the preceding assignments returns false.
Here's some code that illustrates the Integer pool:
public class IntegerPoolTest {
public static void main(String args[]){
Integer i1 = 100;
Integer i2 = 100;
// Comparison of integers from the pool - returns true.
compareInts(i1, i2);
Integer i3 = 130;
Integer i4 = 130;
// Same comparison, but from outside the pool
// (not in the range -128 to 127)
// resulting in false.
compareInts(i3, i4);
Integer i5 = new Integer(100);
Integer i6 = new Integer(100);
// Comparison of Integers created using the 'new' keyword
// results in new instances and '==' comparison leads to false.
compareInts(i5, i6);
}
private static void compareInts(Integer i1, Integer i2){
System.out.println("Comparing Integers " +
i1 + "," + i2 + " results in: " + (i1 == i2) );
}
}
if i write below code(in java):
Integer a =new Integer(5);
Integer b=new Integer(5);
if(a==b){
System.out.println("In ==");
}
if(a.equals(b)){
System.out.println("In equals");
}
My output is: "In equals"
But if i change first and second line to ->
Integer a =5;
Integer b=5;
then my o/p is:
In ==
In equals
So what is difference in creating a Integer object? How it gets created when we do Integer a =5?
Does it mean that a and b object refer to same object, if i create Integer a=5 and creates another object Integer b=5 ?
Integer a = 5; is called autoboxing, compiler converts this expression into actual
Integer a = Integer.valueOf(5);
For small numbers, by default -128 to 127, Integer.valueOf(int) does not create a new instance of Integer but returns a value from its cache. So here
Integer a = 5;
Integer b= 5;
a and b point to the same Object and a == b is true.
In Java, you should never use the new Integer, even though it is valid syntax it is not the best way to declare integers as you have found out. Use instead Integer.valueOf(int) This has multiple advantages.
You are not creating extra objects needlessly. Whenever you use the new operator you are forcing the vm to create a new object which is not needed in most cases. valueOf(int) will return a cached copy. Since Integer objects are immutable this works great. This means that you can use == though in practice you should use a null safe compare like in Apache ObjectUtils
The == operator tests for equality. References are only equal when they refer to the same object in memory. equals method ensures 2 object instances are 'equivalent' to each other.
Integer a = new Integer(5);
Integer b = a;
a == b; // true!
a.equals(b); // true
b = new Integer(5);
a == b; // false
a.equals(b); // true
Primitives are equal whenever their value is the same.
int a = 5; int b = a;
a == b; // true!
for primitive data types like int the equality operator will check if the variables are equal in value
for reference data types like your java.lang.Integer objects, the equality operator will check if the variables reference the same object. In the first case, you have two "new" and separate integer objects, so the references are different
Integer wrapper shares few properties of String class. In that it is immutable and that can be leveraged by using intern() like functionality.
Analyse this:
String a = "foo";
String b = "foo";
String c = new String("foo");
String d = new String("foo");
a == b //true
c == d //false
The reason is when JVM creates a new String object implicitly it reuses existing String object which has the same value "foo", as in case a and b.
In your case JVM implicitly auto-boxes the ints and re-uses existing Integer object. Integer.valueOf() can be used explicitly to re-use existing objects if available.
I believe when you create using new operator it creates object reference. In first case, there are two object references and they are not same but their value is same. That is not the situation in second case.
I have integers that are supposed to be equal (and I verify it by output). But in my if condition Java does not see these variables to have the same value.
I have the following code:
if (pay[0]==point[0] && pay[1]==point[1]) {
game.log.fine(">>>>>> the same");
} else {
game.log.fine(">>>>>> different");
}
game.log.fine("Compare:" + pay[0] + "," + pay[1] + " -> " + point[0] + "," + point[1]);
And it produce the following output:
FINE: >>>>>> different
FINE: Compare:: 60,145 -> 60,145
Probably I have to add that point is defined like that:
Integer[] point = new Integer[2];
and pay us taken from the loop-constructor:
for (Integer[] pay : payoffs2exchanges.keySet())
So, these two variables both have the integer type.
Check out this article: Boxed values and equality
When comparing wrapper types such as Integers, Longs or Booleans using == or !=, you're comparing them as references, not as values.
If two variables point at different objects, they will not == each other, even if the objects represent the same value.
Example: Comparing different Integer objects using == and !=.
Integer i = new Integer(10);
Integer j = new Integer(10);
System.out.println(i == j); // false
System.out.println(i != j); // true
The solution is to compare the values using .equals()…
Example: Compare objects using .equals(…)
Integer i = new Integer(10);
Integer j = new Integer(10);
System.out.println(i.equals(j)); // true
…or to unbox the operands explicitly.
Example: Force unboxing by casting:
Integer i = new Integer(10);
Integer j = new Integer(10);
System.out.println((int) i == (int) j); // true
References / further reading
Java: Boxed values and equality
Java: Primitives vs Objects and References
Java: Wrapper Types
Java: Autoboxing and unboxing
If they were simple int types, it would work.
For Integer use .intValue() or compareTo(Object other) or equals(Object other) in your comparison.
In java numeric values within range of -128 to 127 are cached so if you try to compare
Integer i=12 ;
Integer j=12 ; // j is pointing to same object as i do.
if(i==j)
print "true";
this would work, but if you try with numbers out of the above give range they need to be compared with equals method for value comparison because "==" will check if both are same object not same value.
There are two types to distinguish here:
int, the primitive integer type which you use most of the time, but is not an object type
Integer, an object wrapper around an int which can be used to use integers in APIs that require objects
when you try to compare two objects (and an Integer is an object, not a variable) the result will always be that they're not equal,
in your case you should compare fields of the objects (in this case intValue)
try declaring int variables instead of Integer objects, it will help
The condition at
pay[0]==point[0]
expression, uses the equality operator == to compare a reference
Integer pay[0]
for equality with the a reference
Integer point[0]
In general, when primitive-type values (such as int, ...) are compared with == , the result is true if both values are identical. When references (such as Integer, String, ...) are compared with == , the result is true if both references refer to the same object in memory.
To compare the actual contents (or state information) of objects for equality, a method must be invoked.
Thus, with this
Integer[] point = new Integer[2];
expression you create a new object that has got new reference and assign it to point variable.
For example:
int a = 1;
int b = 1;
Integer c = 1;
Integer d = 1;
Integer e = new Integer(1);
To compare a with b use:
a == b
because both of them are primitive-type values.
To compare a with c use:
a == c
because of auto-boxing feature.
for compare c with e use:
c.equals(e)
because of new reference in e variable.
for compare c with d it is better and safe to use:
c.equals(d)
because of:
As you know, the == operator, applied to wrapper objects, only tests whether the objects have identical memory locations. The following comparison would therefore probably fail:
Integer a = 1000;
Integer b = 1000;
if (a == b) . . .
However, a Java implementation may, if it chooses, wrap commonly occurring values into identical objects, and thus the comparison might succeed. This ambiguity is not what you want. The remedy is to call the equals method when comparing wrapper objects.