I have a basic doubt, we can increment reference of Integer class object, while not reference of any other class(D class has 1 Integer data member and one parameterized constructor).
Integer x=new Integer(10); x++;
D d=new D(10); d++;
here both x and d are reference still we are able to increment Integer reference while not any other reference. I am missing something very basic, please help me out.
Integer x=new Integer(10); x++;
Here java compiler will do the un-boxing of the object, which converting the Integer object into primitive int. Then the increment operation will be performed. This is only defied for java primitives, the auto-boxing is converting primitive to wrapper object and reverse of it is call un-boxing. This is what is happening here.
The auto-boxing or un-boxing which is an automatic conversion, it is defined for java primitives only. So it can not be performed on other objects. Remember, the object are not just memory references like C or C++ that we can increment them.
This is because of Autoboxing for wrapper classes which was introduced from jdk 1.5. Internally java will convert the integer reference to int and increment it. You can't increment the objects.
Java has no way of overloading operators for arbitrary classes. Thus, D d; d++ is just not available. You can do:
public class D {
private int value;
public D( int v ){
value = v;
}
public void increment(){
value++;
}
}
Now you can do
D d = new D(10);
d.increment();
and d's value will be 11.
According to Oracle Java Documentation, The automatic conversion of primitive data types into its equivalent wrapper type is known as autoboxing and opposite operation is known as unboxing. Whenever we use object of wrapper class in an expression, automatic unboxing and autoboxing is done by JVM.
Integer object;
object = 100; //Autoboxing of int
++object;
When we perform an increment operation on Integer object, it is first unboxed, then incremented and then again reboxed into Integer type object. You can also take a look at How java auto boxing/unboxing works?
Related
I am wondering what is the internal Java behaviour for the next snippet:
Long a = 123L;
long b = 123;
System.out.println("a equals b?: " + (a == b));
The result is true although comparing two Long objects would be false (because it compares their reference). It is Java converting Long object into its primitive value because detects == operator against another primitive object?
It is Java converting Long object into its primitive value because detects == operator against another primitive object?
Yes. One of the operands is a primitive type and the other operand is convertible to a primitive by unboxing.
JLS section 15.21.1 says (emphasis mine):
If the operands of an equality operator are both of numeric type, or one is of numeric type and the other is convertible (§5.1.8) to numeric type, binary numeric promotion is performed on the operands (§5.6.2).
Note that binary numeric promotion performs value set conversion (§5.1.13) and may perform unboxing conversion (§5.1.8).
Also, it is important to note that reference equality is only performed if both operands are objects. JLS section 15.21.3 says:
If the operands of an equality operator are both of either reference type or the null type, then the operation is object equality.
From the Java Tutorials:
Autoboxing and Unboxing
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));
Converting a primitive value (an int, for example) into an object of
the corresponding wrapper class (Integer) is called autoboxing. The
Java compiler applies autoboxing when a primitive value is:
Passed as a parameter to a method that expects an object of the
corresponding wrapper class.
Assigned to a variable of the
corresponding wrapper class.
Consider the following method:
public static int sumEven(List li) {
int sum = 0;
for (Integer i: li)
if (i % 2 == 0)
sum += i;
return sum;
}
Because the remainder (%) and unary plus (+=) operators do not apply
to Integer objects, you may wonder why the Java compiler compiles the
method without issuing any errors. The compiler does not generate an
error because it invokes the intValue method to convert an Integer to
an int at runtime:
public static int sumEven(List<Integer> li) {
int sum = 0;
for (Integer i : li)
if (i.intValue() % 2 == 0)
sum += i.intValue();
return sum;
}
Converting an object of a wrapper type (Integer) to its corresponding
primitive (int) value is called unboxing. The Java compiler applies
unboxing when an object of a wrapper class is:
Passed as a parameter to a method that expects a value of the corresponding primitive type.
Assigned to a variable of the corresponding primitive type.
Can someone please explain how/why is this allowed in Java?
public class Test {
private int text;
public Integer getText() {
return text;
}
I am basically having the wrapper class as the return type, while I am infact returning a primitive.
Because Java supports Autoboxing and Unboxing in versions 5 and up. That is an example of the former, but the later is equally important (and the reverse conversion). Per the link,
Autoboxing is the automatic conversion that the Java compiler makes between the primitive types and their corresponding object wrapper classes.
Consider the following code: (From: http://docs.oracle.com/javase/tutorial/java/data/autoboxing.html)
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));
Converting a primitive value (an int, for example) into an object of the corresponding wrapper class (Integer) is called autoboxing. The Java compiler applies autoboxing when a primitive value is:
Passed as a parameter to a method that expects an object of the corresponding wrapper class.
Assigned to a variable of the corresponding wrapper class.
From javadocs : Since java 5 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.
The java compiler applies autoboxing usually when -
Passed as a parameter to a method that expects an object of the
corresponding wrapper class.
Assigned to a variable of the corresponding wrapper class.
For the sake of performance, not everything in Java is an object. There are also primitives, such as int, long, float, double, etc.
For example java.lang.Integer class :-
It wraps an int.
Has two static final fields of type int: MIN_VALUE and MAX_VALUE.
MIN_VALUE contains the minimum possible value for an int (-2^31) and MAX_VALUE the maximum possible value for an int (2^31 - 1).
It also has two constructors - public Integer (int value) & public Integer (String value).
Integer has the no-arg byteValue, doubleValue, floatValue, intValue, longValue, and shortValue methods that convert the wrapped value to a byte, double, float, int, long, and short, respectively.
In addition, the toString method converts the value to a String.
Static methods parse a String to an int (parseInt) and convert an int to a String (toString).
class AutoBox {
public static void main(String args[]) {
// autobox an int
Integer a = 100;
// auto-unbox
int b = a;
System.out.println(b + " " + a); // displays 100 100
}
}
This feature has been added in Java 5.
text gets converted automatically into Integer by compiler. So basically its a syntactic sugar that can shorten your code (otherwise you would do conversions back and forth by yourself). Of course it has its price and if this happens a lot (I mean really a lot, big loops, frequent invocations and so forth), it can become a performance issue, so when using it, just remember the it happens under the hood and you'll be fine.
Integer.valueOf(text)
is called under the hood
The feature is called autoboxing btw
Hope this helps
class Test{
public static void main(String[] args){
int a = 1;
int b = 5;
Integer c = new Integer(1);
Integer d = 5; //autoboxing at work
System.out.println(c.compareTo(d));
System.out.println(a.compareTo(b));
}
}
Why doesn't a.compareTo(b) compile (int cannot be dereferenced)? I know that compareTo requires objects, but I thought autoboxing would automatically make an int an Integer when necessary. Why doesn't autoboxing occur in this case? And what other cases will it not occur?
From the Oracle tutorial on Autoboxing, the two cases where boxing will occur are, when primitives are:
Passed as a parameter to a method that expects an object of the corresponding wrapper class.
Assigned to a variable of the corresponding wrapper class.
The expression being evaluated in your example (a.compareTo(d)) does not fall in any one of those scenarios.
Theres some interesting additional information in the JCP proposal for autoboxing, describing the mechanics and rules for assignment conversion, method invocation conversion, and casting conversion.
Considering Java. How are these 2 different and why?
public void languageChecks() {
Integer a = 5;
Integer b = new Integer(5);
change(a); // a doesn't get incremented. value is 5
change(b); // b does. value is now 6
}
public void change(Integer a) {
a++;
}
The only difference is that
Integer b = new Integer(5);
guarantees a new object is created. The first will use an instance from a cache (see Integer.valueOf()).
Both are immutable and the references to both are passed by value (as is everything in Java). So change() has no effect on either.
I'd always been taught a++ was just shorthand for a = a + 1 in which case a local variable is created named a and immediately thrown away when the method returns. There're no methods on Integer that change the value (it's immutable), and likewise no operations on primitive ints that change their value.
Neither call to change() affects the values passed in, because of auto-boxing/unboxing.
public void change(Integer a) {
// This unboxes 'a' into an int, increments it and throws it away
a++;
}
The above code seems to imply that a++ changes the value of a, since it's an object, not a primitive. However, ++ is not overloaded by Integer, so it unboxes it to be able to apply the ++operator on its int. To me the compiler shouldn't allow this.
is declaring/initializing primitives the same as creating new objects? from what i know when we create primitives, we also creating wrapper classes for them. im implementing on java btw.
No, assigning primitive values does not create any objects.
What you might be referring to is the fact that primitive values can be auto-boxed into the corresponding wrappers, when they are used in a context where a reference type (a.k.a "an object") is required:
int i = 13; // this line does not create an object
Integer i2 = i; // at this line 13 is auto-boxed into an Integer object
char c = 'x'; // again: no object created:
List<Character> l = new ArrayList<Character>();
l.add(c); // c is auto-boxed into a Character object
Also, I'll try to describe the difference between declare and initialize:
int i; // an int-variable is declared
int j = 0; // an int-variable is declared and initialized
i = 1; // an int-variable is assigned a value, this is *not* initialization
A variable is "declared" when it is created for the first time (i.e. you specify the type and name of the variable). It is initialized when it's assigned a value during declaration.
No, declaring and initializing a primitive variable does not create an object. Let's take a class with two integer values - one using the wrapper type and one not.
public class Foo
{
private int primitive = 10;
private Integer wrapper = new Integer(10);
}
The value of the primitive variable is just the number 10. The value of the wrapper variable is a reference to an Integer object which in turn contains the number 10. So an instance of Foo would keep state for the number in primitive and the reference in wrapper.
There are wrapper classes for all primitive types in Java, but you don't use them automatically.
Creating a primitive DOES NOT also create a wrapper class for them.
As for your original question: Declaring/initializing a primitive will create it on the stack, whereas declaring an object will allocate a variable to hold a reference to an object. Initializing the object will allocate it on the heap.
Answer: No.
Check this out: http://java.sun.com/docs/books/tutorial/java/nutsandbolts/datatypes.html
No. Primitives are not objects in java.