What is the difference between Integer.class, Integer.TYPE and int.class?
acc to me
Integer.class is a reference of Integer (Wrapper) Class object
but what is then int.class as int is not a class, it's a primitive type. And what does Integer.TYPE refer to?
Integer.class is, as you say, a reference to the Class object for the Integer type.
int.class is, similarity, a reference to the Class object for the int type. You're right that this doesn't sound right; the primitives all have a Class object as a special case. It's useful for reflection, if you want to tell the difference between foo(Integer value) and foo(int value).
Integer.TYPE (not Integer.type, mind you) is just a shortcut for int.class.
You can get a sense of this with a simple program:
public class IntClasses {
public static void main(String[] args) {
Class<Integer> a = int.class;
Class<Integer> b = Integer.TYPE;
Class<Integer> c = Integer.class;
System.out.println(System.identityHashCode(a));
System.out.println(System.identityHashCode(b));
System.out.println(System.identityHashCode(c));
}
}
Example output (it'll be different each time, but the first two will always be the same, and the third will virtually always be different):
366712642
366712642
1829164700
From java.lang.Class.isPrimitive API
There are nine predefined Class objects to represent the eight primitive types and void. These are created by the Java Virtual Machine, and have the same names as the primitive types that they represent, namely boolean, byte, char, short, int, long, float, and double.
These objects may only be accessed via the following public static final variables java.lang.Boolean.TYPE, java.lang.Integer.TYPE etc
Java handles primitive types versus class types in a schizophrenic way by defining two types for each primitive.
For instance int is the primitive type and Integer the class type. When you use generics, you are forced to use a non-primitive type so ArrayList<Integer> is allowed but ArrayList<int> not.
Since you sometimes want to perform reflection, this duality results in two classes (how else can you inspect a method public int foo ();).
Say you have a class:
public class Foo {
private Integer value;
public int value1 () {
return value;
}
public Integer value2 () {
return value;
}
}
The two methods will not always return the same value, since value2() can return null and value1() will throw a runtime error.
In plain terms :
int -- > Are primitives..for simple math operations. You cannot add them
to a collection.
Integer --> Are objects in themselves.. are wrappers to ints. i.e,
they can be used with collections (as they are objects). They are
collected as normal objects by the GC.
EDIT :
public static void main(String[] args) {
int i = 5;
System.out.println(int.class);
Integer i1 = new Integer(5);
System.out.println(Integer.TYPE);
}
O/P : int
int
So, basically, both return an int. Integer.TYPE just returns the primitive type of the Integer class. It is true for any wrapper class
To me, the easiest way to understand int.class and Integer.class is to stop thinking Integer is a wrapper (which kind of implies it is "special"). In fact, it is easier, and probably more appropriate, to think of int as a special case.
Integer is just a normal Java class, nothing different from e.g. String. It derives from Object, operates like Object, At runtime you can create an instance of Integer, the instance takes an object-like memory layout, e.g. with a pointer at the beginning pointing to the Integer.class, which is what enables the polymorphic runtime behavior of java.
There is really nothing special yet about Integer. if you imagine a Java without boolean, int, long these primitives, but only with Integer, Boolean, Long etc, the type system is actually very consistent.
Conceptually, you can think of int as a special class introduced later for performance reasons. Initially, it has nothing to do with Integer. At the time when Java was created, maintaining an object-like memory layout for plain numbers is very expensive for arithmetic heavy programs. And most of the arithmetic operations do not even involve polymorphic dynamic dispatching at all. E.g. it is a lot less common to invoke methods such as toString on a number.
int is special in the sense that it is a class whose "instances" are laid out in memory with the common object structure stripped off - just four consecutive bytes with no extra meta data.
As a result, you cannot do 123.getClass() because the runtime memory layout of int 123 does not have a class pointer. int.class does exist, it is completely unrelated to Integer.class (yet). In a sense, int is more similar to Void, as in Void.class does exist, but you can never have object o where o.class == Void.class.
Java could just settle here, int is int, Integer is Integer. But what people realize is that although less common, it is still very useful to be able treat int as a normal Java object otherwise you will always have to maintain two sets of your methods at least - one that takes normal objects, and another one that takes primitives - even though performance is not your concern. In these scenarios, int is actually behaving like Integer. Java allows this conversion to automatically happen through the processes of auto-boxing, hence effectively making int and Integer related.
But int is still int, and Integer is still Integer (e.g. int.class != Integer.class). But one extra field is introduced to Integer.class to indicate it relates to int.class, that is Integer.TYPE. So Integer.TYPE == int.class. To me Integer.TYPE is just to capture the notional that Integer and int are related. How to use it, or whether it is even useful, is up to you.
In practice, int and Integer are still separate types where it matters. For example:
void accept (int value);
void accept (Integer value);
are still considered two overloads. Hence when working with reflection, you can use int.class and Integer.class to differentiate between the two.
When not working with reflection or some form of meta programming, because you cannot navigate to int.class from an object, it is relatively rare to see int.class to be used in your code. Sometimes it feels confusing because auto-boxing syntax seems to suggest you should be getting int.class:
int value = 1;
Class c = ((Object)value).getClass();
But that is just an illusion, as the moment you do ((Object)value), you are actually creating a new Integer instance with the same value.
The only time I need to explicitly work with int.class and Integer.class is to build a generics api <T> T getValue(String name, Class<T> type); where I need to differentiate if the api is allowed to return null:
int value = getValue("key", Integer.class); // will potentially throw NPE
int value = getValue("key", int.class); // will never throw NPE
Related
This question already has answers here:
What is the difference between an int and an Integer in Java and C#?
(26 answers)
Closed 2 years ago.
I have seen many times in code that people use int or Integer to declare variable in beans. I know int is datatype and Integer is wrapper class.
My question is, in which condition int or Integer should be used and is there any advantage of either of them over another?
My question is, in which condition int or Integer should be used and is there any advantage of either of them over another?
Well, you should use the reference type Integer whenever you have to. Integer is nothing more than a boxed int. An Object with a single field containing the specified int value.
Consider this example
public class Ex {
int field1;
Integer field2;
public Ex(){}
}
In this case field1 will be initialized with the value 0 while field2 will be initialized with null. Depending on what the fields represent, both approaches might be advantageous. If field1 represents some kind of UUID, would you want it to be initialized with a zero value?
I wouldn't worry too much about the performance implications of Autoboxing. You can still optimize after you get your code running.
For more information take a look at the documentation
https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/lang/Integer.html
You always use int, pretty much.
Integer should rarely be used; it is an intermediate type that the compiler takes care of for you. The one place where Integer is likely to appear is in generics, as int is simply not legal there. Here is an example:
List<Integer> indices = new ArrayList<Integer>();
int v = 10;
indices.add(v);
The above works: It compiles with no errors and does what you think it would (it adds '10' to a list of integer values).
Note that v is of type int, not Integer. That's the correct usage; you could write Integer here and the code works as well, but it wouldn't be particularly idiomatic java. Integer has no advantages over int, only disadvantages; the only time you'd use it, is if int is straight up illegal. Which is why I wrote List<Integer> and not List<int> as the latter is not legal java code (yet - give it a few versions and it may well be legal then, see Project Valhalla).
Note also that the compiler is silently converting your v here; if you look at the compiled code it is as if javac compiled indices.add(Integer.valueOf(v)) here. But that's fine, let the compiler do its thing. As a rule what the compiler emits and what hotspot optimizes are aligned; trust that what javac emits will be relatively efficient given the situation.
int is a primitive type, a value type for number literals.
it is used whenever and wherever you just want to do some basic arithmetical operation;
it is a value type, so it's stored in the Stack area of the memory, hence operations on it are much faster;
whenever it's needed, compiler implicitly and automatically casts back and forth (a.k.a Boxing and Unboxing) from int to Integer and vice versa;
Integer is a Class, which is a reference type and you instantiate an Object of that type.
you create an object of that class, which means, that you also have some methods and operations on that object;
any time you do some arithmetic operation on the instance of Integer, under the hood, it's still implemented by int primitives, and it's just wrapped into box/container;
it is a reference type / object, which is very important, as you can Serialize or Deserialize it;
it also has some very useful utility factory methods, like Integer.valueOf(..) for example, to parse the String as an integer;
it can be well used into declarations of the generic types and it, as a class, supports the hierarchy as well. For instance, it extends Number, and you can make use of this;
it is stored in the Heap area of the memory.
int is a primitive and Integer is an object .
From an memory footprint point of view , primitive consume less memory than object and also primitives are immutable (since they use pass by value ) .
Here is a good article on when to use what :
https://www.baeldung.com/java-primitives-vs-objects
What is the difference between Integer.class, int.class and Integer.TYPE? I am kind of confused between the three.
Whether i is an int or an Integer:
Integer.class.isInstance(i) returns true
int.class.isInstance(i) returns false
Integer.TYPE.isInstance(i) returns false
Let's understand isInstance.
public boolean isInstance(Object obj)
Determines if the specified Object is assignment-compatible with the object represented by this Class.
It takes an Object as an argument, not a primitive type. Any int passed in will be boxed as an Integer so it can be passed to the method.
Integer.class is a class literal, and used with a reference type, that is a reference to the Class object for that class.
The Integer object that this method sees certainly is an instance of the Integer class, so Integer.class.isInstance(i) returns true.
However, int.class and Integer.TYPE each represent the int primitive type, not the Integer class, and an Integer is not an int, despite the fact that Java usually uses them interchangeably, due to boxing and unboxing. This explains the two false outputs from int.class.isInstance(i) and Integer.TYPE.isInstance(i).
Class literal history
According to the Javadocs for Integer, Integer.TYPE has been around since version 1.1.
But class literals have also been around since 1.1.
Java APIs which require class objects as method arguments are much easier to use when the class literal syntax is available.
The class literals made it tremendously easier to use reflection, which was also new in Java 1.1.
It is unclear why Integer.TYPE exists if int.class represents the int primitive type consistently with reference class literals. They are equivalent and == yields true when comparing them.
I read this post: Is int an object in Java?.
In the post it is argued that int is not inherited from Object. If so is the case, then why does the code below compile without any error? Given that int is not Object and the signature of format() method is public static String format(String format, Object... args) as shown in documentation: javadoc for String!
public class Testing {
public static void main(String[] args) {
int integer = 7;
String str = String.format("%03d", integer);
System.out.println(str);
}
}
I have also read about "Autoboxing". What does this exactly mean? Are all the primitives replaced by appropriate Object's before compilation? If so, then is there any advantage of memory usage while using large array of int's (int[]) over Integer's (Integer[])? Similar arguments follow for double's etc.
Any insights are welcome.
It is caused by Autoboxing.
Here is a small snippet from the linked Java documentation that explains it better than I could:
Autoboxing is the automatic conversion that the Java compiler makes
between the primitive types and their corresponding object wrapper
classes. For example, converting an int to an Integer, a double to a
Double, and so on. If the conversion goes the other way, this is
called unboxing.
Here is the simplest example of autoboxing:
Character ch = 'a';
The rest of the examples in this section use generics. If you are not
yet familiar with the syntax of generics, see the Generics (Updated)
lesson.
Consider the following code:
List<Integer> li = new ArrayList<>();
for (int i = 1; i < 50; i += 2)
li.add(i);
Although you add the int values as primitive types, rather than
Integer objects, to li, the code compiles. Because li is a list of
Integer objects, not a list of int values, you may wonder why the Java
compiler does not issue a compile-time error. The compiler does not
generate an error because it creates an Integer object from i and adds
the object to li. Thus, the compiler converts the previous code to the
following at runtime:
List<Integer> li = new ArrayList<>();
for (int i = 1; i < 50; i += 2)
li.add(Integer.valueOf(i));
When calling String.format("%d",myInt), myInt is automatically (and implicitly) wrapped in an Integer instance, which extends Object, therefore it compiles.
Concerning the arrays, the conversion from primitiveType[] to WrapperClass[] is not automatic for some reason. If you try to use an array of a primitive type where an array of the wrapper class is expected, it will result in a compile error.
Using Integer creates an overhead compared to using int because you need to assign and store references. However, this overhead is limited when using Integer values between -128 and 127 because these values are pooled (which means that all instances of Integer wrapping a value in this in interval point to a unic reference).
Autoboxing is a help from the compiler, which automatically compiles something like
foo(i);
into
foo(Integer.valueOf(i));
when foo() takes an argument of type Object and you pass it a primitive type (int, in this case). It just makes the code easier to type and read.
And that's what happens here. The String.format() method expects objects as argument. You're passing it a primitive type, so the compiler autoboxes it to an Integer for you.
Just trying to understand the behavior of java.lang.Object class.
public class TypeCheck{
static void printMethod(Object obj)
{
System.out.println(obj);
}
public static void main(String[] args)
{
Object obj = new Object();
Integer intObj = new Integer(12);
String stringObj = new String("Hello");
printMethod(obj); //---> java.lang.Object#78b5f53a
printMethod(intObj); //---> 12
printMethod(stringObj); // ---> Hello
}
}
My questions are:
When should we use Object class?
But, when I only pass printMethod(intObj) and in printMethod(Object obj) I do an addition: System.out.println(obj+1) it does not work. If obj recognizes that it's an Integer, why can't I do operations on it?
Exception: TypeCheck.java:5: operator + cannot be applied to java.lang.Object,int System.out.println(obj+1);
Now, if I do this:
public class TypeCheck
{
static void printMethod(Object obj)
{
System.out.println(obj.getClass().getName());
}
public static void main(String[] args)
{
int i = 666;
printMethod(i);
}
}
It returns java.lang.Integer, but it's defined as int (primitive type). Why does Java convert a primitive type to it's wrapper class when passed to an Object.
Almost never. You should use as specific an object you can.
That should work. See below
Because primitives, i.e. int, are not objects. Therefore it needs to be made an object (boxed) to be passed as one.
EDIT in response to question edit:
(Number 2)
Inside of printMethod, the type of obj is not int, it is Object. You can't add objects! You would need to cast it back to Integer or int before you could perform such operations on it. What would the + do if you passed an instance of TypeCheck into TypeCheck.printMethod?
You rarely need to use Object itself. Object is a placeholder that contains default implementations of things like toString(), hashCode() and so on. You don't need to explicitly extend it to get these, though - every class inherits them automatically.
Java is a statically typed language, which means that type checking takes place at compile time to ensure all parameters are of the correct type, methods exist on the (declared) type of the objects they're being called on, and so on. The compiler doesn't know that you might pass an Integer into printMethod, and obj+1 makes no sense when obj is a string or a bare Object. You could cast it to an Integer, but of course this will break (ClassCastException) when you pass in a non-Integer.
Primitive types are not objects in Java. When the Java compiler sees code that uses a primitive type where an object is required:
int i = 5;
printMethod(i);
it compiles as if it were:
int i = 5;
printMethod(new Integer(i));
This is known as autoboxing and was introduced in Java 5 so that primitive types could be stored inside collections.
An addition (like in your second question) becomes something like:
// ... assume obj has been cast to an Integer ...
System.out.println(obj.intValue() + 1);
which of course makes no sense if obj were actually an Object, since it wouldn't have an integer value to begin with!
Question 1
When you have a list/array of different objects of different classes and you need to access each object. (I know this is bad and we should use Generics, but that is beside the point) Object is the superclass of all class and you can use it to reference any other objects.
Question 2
Cast it to Integer before you can perform addition as the addition operation belongs to the Integer class and not the Object class.
When you print a object, the toString() method is called. The implementation of toString() in Object class returns the pointer and the implementation of toString() in Integer returns the value of the integer. Depending on the type of the object, the corresponding toString() method will be called although the instance is the same.
Question 3
The other concept you have to understand is AutoBoxing. Java automatically converts your primitive int to Integer.
I believe you will rarely use Object directly. I mean you will mostly never instantiate it. The purpose of Object is to be a common parent to all other objects (all classes inherit from the Object class). So Object will define methods that all objects in Java are going to have, like toString. You will sometimes use an Object pointer to hold a class of unknown identity but, most of the time, you can find a closer parent to the candidate classes which can be held by this pointer.
When you pass intObj (which is an Integer, but can pass as an Object since Integer inherits from Object), it is not recognized by the method as an Integer. If it prints a number, it's because the toString method in the class Integer was overrided to print the value it represents, instead of the name of the class and its address. See it this way too... You can't do operations on the Integer in the printMethod, because printMethod doesn't know if it is going to be a raw Object, an Integer, or anything else. If you cast the Object as an Integer, it should get unboxed and the operation will succeed.
It's called autoboxing. Boxing is putting a primitive in an object which basicly only holds the primitive. So when you pass the int to a method which is waiting for an object, the int it boxed in an Integer and then passed to the method.
I had a function in Java with method signature
public void myMethod (int someInt, String someString)
in my abstract class and I had over-ridden it with method
public void myMethod (Integer someInt, String someString)
The over ride does not work. Is this an inconsistency ? I thought auto-boxing applied to method signature over-ride as well.
int and Integer are two different types. Autoboxing blurs the distinction at the source code level for the convenience of programmers, but does not change the fact that they are in fact two very different types.
As such, you can not #Override a method that takes an int with one that takes an Integer and vice versa.
Note that you should probably think twice before declaring a method to take an Integer instead of an int. Here's an excerpt from Effective Java 2nd Edition, Item 49: Prefer primitives to boxed primitives:
In summary, use primitives in preference to boxed primitive whenever you have the choice. Primitive types are simpler and faster. If you must use boxed primitives, be careful! Autoboxing reduces the verbosity, but not the danger, of using boxed primitives. When your program compares two boxed primitives with the == operator, it does an identity comparison, which is almost certainly not what you want. When your program does mixed-type computations involving boxed and unboxed primitives, it does unboxing, and when your program does unboxing, it can throw NullPointerException. Finally, when your program boxes primitive values, it can result in costly and unnecessary object creations.
There are places where you have no choice but to use boxed primitives, e.g. generics, but otherwise you should seriously consider if a decision to use boxed primitives is justified.
See also
Java Language Guide/Autoboxing
Related questions
What is the difference between an int and an Integer in Java/C#?
Is it guaranteed that new Integer(i) == i in Java? (YES!!! The box is unboxed, not other way around!)
Why does int num = Integer.getInteger("123") throw NullPointerException? (!!!)
Why null == 0 throws NullPointerException in Java?
No, these two signatures define two different methods.
They are absolutely not overridden but overload since the parameters are different. And JVM will choose the method to launch base on this:
widen - boxing - var args...
For example, you have three methods
void add(long n) {} // call this method 1
void add(int... n) {} // call this method 2
void add(Integer n) {} // call this method 3
and when you invoke:
int a = 5;
add(a);
the method 1 will be invoked.
The reason the override doesn't work is because Integerand int are two different things. Integer is an object, whereas int is a primitive type. Java does the implicit typecasting for you. For example:
int myInteger = new Integer(5);
Will create a primitive int type called myInteger with a value of 5. As the Javadoc says,
"The Integer class wraps a value of
the primitive type int in an
object."
You are correct that this scenario will not work because Java provides the functionality of the autoboxing, so at runtime JVM can not decide which method to call because it can call both method as both method fits for the argument type. So I think it will give an error or will choose either method randomly..... Just run it and see.....
int and Integer are two different types in JAVA.Although autoboxing specifies the distinction at the source code level, but does not change the eternal fact that they are in fact two very different types.