I don't get why java doesn't do the widening and then autoboxing.
Integer i = (short) 10;
I would think the following would take place:
Narrowing conversion first from 10 to short.
short would then widen to int.
int would then autobox to Integer.
Instead it's a compilation error.
Example 2:
Short x = 10;
Integer y = x;
This fail too.
According to the JLS, Section 5.2, which deals with assignment conversion:
Assignment contexts allow the use of one of the following:
an identity conversion (§5.1.1)
a widening primitive conversion (§5.1.2)
a widening reference conversion (§5.1.5)
a boxing conversion (§5.1.7) optionally followed by a widening
reference conversion
an unboxing conversion (§5.1.8) optionally followed by a widening
primitive conversion.
It is unable to apply two conversions at once (widening primitive conversion and boxing conversion); only one conversion can apply here, so it has to result in an error.
The solution would be to cast the short back to an int (a casting conversion), which would allow the assignment conversion to be a boxing conversion:
Integer i = (int) (short) 10;
(Or here, don't cast it to short in the first place.)
Integer i = 10;
What's going on here is a casting conversion from int to short, and then an attempted assignment conversion from short to Integer.
Assignment conversion (§5.2) allowed for boxing and then widening, but not widening and then boxing.
Assignment contexts allow the use of one of the following:
an identity conversion (§5.1.1)
a widening primitive conversion (§5.1.2)
a widening reference conversion (§5.1.5)
a boxing conversion (§5.1.7) optionally followed by a widening
reference conversion
an unboxing conversion (§5.1.8) optionally followed by a widening
primitive conversion.
IN java it follows sequence as "autoBoxing and then widening" only no matter if you are doing
this :
int x =5;
Object obj = x;
Or
this:
int x = 5;
Long l = x;
Widening occurs only when there is a is-a relationship.
So while applying above sequence first case is very much valid for compiler, because int would be autobox to Integer and then assigned to Object, i.e. widening (i.e. autobox first and then widening), being in Is-A relationship. But in second case if int x is autbox to Integer and the assign to Long it will not be allowed, being not is-a relationship, hence throws compilation error.
Could simulate the similar use case using overloading with respect to autoboxing and widening.
public static void m(short s) {
System.out.println("widening");
}
public static void m(Integer i) {
System.out.println("Autoboxing");
}
public static void main(String[] args) {
short x = 10;
m(x);
}
Output: widening
Hence, in nut shell we could say, Widening dominates Autoboxing in java
Related
public class Yikes1 {
public static void go(Long n) {
System.out.println("Long "); // printed
}
public static void go(Short n) {
System.out.println("Short "); // don't know why isn't this printed
}
public static void go(int n) {
System.out.println("int "); // printed
}
public static void main(String [] args) {
short y = 6;
long z = 7;
go(y);
go(z);
}
}
How does the short value got converted to int before it prints the output?
I thought widening works only when the dataype-short is not available so it looks for next available data type which is int in this case but however short datatype is defined here so how come the automatic promotion happening?
The binding sequence works as follows:
Exact match (Ex. int--> int)
Promotion (Ex. int --> long)
Autoboxing/Unboxing (Ex. int --> Integer)
Varags (Ex. int --> int...)
Okay, when there's no method, which accepts short, there's 2 options: autobox it into Short or cast to integer. JLS states, that the second option in prefered:
Method invocation contexts allow the use of one of the following:
an identity conversion (§5.1.1)
a widening primitive conversion (§5.1.2)
a widening reference conversion (§5.1.5)
a boxing conversion (§5.1.7) optionally followed by widening reference
conversion
an unboxing conversion (§5.1.8) optionally followed by a widening
primitive conversion.
What you expect here is a boxing conversion, but what you get is a widening primitive conversion.
You can read more about boxing here to correctly understand the relation between short and Short.
Consider the following snippet:
Integer Foo = 2;
int foo = 1;
boolean b = Foo < foo;
is < done using int or Integer? What about ==?
For all the relational operators (including therefore < and ==), if one type is the boxed analogue of the other, then the boxed type is converted to the unboxed form.
So your code is equivalent to Foo.intValue() < foo;. This is deeper than you might think: your Foo < foo will throw a NullPointerException if Foo is null.
According to JLS, 15.20.1
The type of each of the operands of a numerical comparison operator must be a
type that is convertible (§5.1.8) to a primitive numeric type, or a compile-time error occurs. Binary numeric promotion is performed on the operands (§5.6.2).
Further, 5.6.2 states that
If any operand is of a reference type, it is subjected to unboxing conversion
This explains what is happening in your program: the Integer object is unboxed before the comparison is performed.
They will be done using int due to Autoboxing and Unboxing.
The Wrapper types for primitive types in java does automatic "type casting" ( or autoboxing / unboxing) from Object to compatible primitive type. so Integer will be converted to int before passing it to comparison operators or arithmetic operators like < , > , == , = , + and - etc.
public class Yikes1 {
public static void go(Long n) {
System.out.println("Long "); // printed
}
public static void go(Short n) {
System.out.println("Short "); // don't know why isn't this printed
}
public static void go(int n) {
System.out.println("int "); // printed
}
public static void main(String [] args) {
short y = 6;
long z = 7;
go(y);
go(z);
}
}
How does the short value got converted to int before it prints the output?
I thought widening works only when the dataype-short is not available so it looks for next available data type which is int in this case but however short datatype is defined here so how come the automatic promotion happening?
The binding sequence works as follows:
Exact match (Ex. int--> int)
Promotion (Ex. int --> long)
Autoboxing/Unboxing (Ex. int --> Integer)
Varags (Ex. int --> int...)
Okay, when there's no method, which accepts short, there's 2 options: autobox it into Short or cast to integer. JLS states, that the second option in prefered:
Method invocation contexts allow the use of one of the following:
an identity conversion (§5.1.1)
a widening primitive conversion (§5.1.2)
a widening reference conversion (§5.1.5)
a boxing conversion (§5.1.7) optionally followed by widening reference
conversion
an unboxing conversion (§5.1.8) optionally followed by a widening
primitive conversion.
What you expect here is a boxing conversion, but what you get is a widening primitive conversion.
You can read more about boxing here to correctly understand the relation between short and Short.
Why is it possible to pass a primitive to a method that takes an object? Is the primitive turned into an object? like int = Integer and boolean = Boolean?
I can call the following function:
hash(41, 0);
public static int hash(int seed, Object object)
{
int result = seed;
if(object == null)
{
return hash(result, 0);
}
else if(!isArray(object))
{
result = hash(result, object.hashCode());
}
else
{
int length = Array.getLength(object);
for(int index = 0; index < length; ++index)
{
Object item = Array.get(object, index);
// prevent looping if item in array references the array itself
if(!(item == object))
{
result = hash(result, item);
}
}
}
return result;
}
Yes, this is called a boxing conversion. The int value is "boxed" into an Integer, which is an Object. It has been available in Java since 1.5.
The JLS, Section 5.1.7 lists available boxing conversions:
Boxing conversion converts expressions of primitive type to
corresponding expressions of reference type. Specifically, the
following nine conversions are called the boxing conversions:
From type boolean to type Boolean
From type byte to type Byte
From type short to type Short
From type char to type Character
From type int to type Integer
From type long to type Long
From type float to type Float
From type double to type Double
From the null type to the null type
Additionally, the boxing conversion is allowed during method invocation conversion, which is really what's going on here. The value is being converted to another type because the int 0 is being passed to a method that expects an Object. The JLS, Section 5.3, lists boxing conversion as one method of method invocation conversion:
Method invocation contexts allow the use of one of the following:
an identity conversion (§5.1.1)
a widening primitive conversion (§5.1.2)
a widening reference conversion (§5.1.5)
a boxing conversion (§5.1.7) optionally followed by widening reference
conversion
an unboxing conversion (§5.1.8) optionally followed by a widening
primitive conversion.
Yes you can. Something called autoboxing/unboxing is done by the compiler automatically. Here is an excerpt from the docs.
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.
int i = 1;
Integer boxI = i; // Autoboxing is performed automatically by the compiler
Integer ii = 1;
int i = ii; // Auto(un)boxing is performed automatically by the compiler
Yes, primitive is converted to Object and vice versa, and this concept is called boxing and unboxing. In newer versions of java this is done automatically, hence called, auto-boxing and auto-unboxing.
Oracle doc for Boxing and UnBoxing
Each primitive has corresponding Wrapper class.
int -> Integer
boolean -> Boolean
char -> Character
and so on.
Is type casting possible between wrapper classes in Java?
The code here is what I tried:
public class Test
{
public static void main(String[] args)
{
Double d = 100.04;
Long l = (long) d.doubleValue(); //explicit type casting required
int i = (int) l; //explicit type casting required
System.out.println("Double value " + d);
System.out.println("Long value " + l);
System.out.println("Int value " + i);
}
}
Why isn't Long casted to int in this program?
Long l = (long)d.doubleValue(); //explicit type casting required
int i = (int)l; //Not valid, because here you are casting a Wrapper Long
In above line, casting of l is not possible to int because Wrapper classes are only by Autoboxing and Unboxing. Refer: https://docs.oracle.com/javase/tutorial/java/data/autoboxing.html.
Wrapper class is casted only to its corresponding primitive type.
long l = (long)d.doubleValue(); //explicit type casting required
int i = (int)l; //valid, because here you are casting primitive to primitive
In above lines casting of l is possible to int because long is primitive type and int is also a primitive type. Here primitive narrowing casting will take place as casting from long to int may result in loss of some precision.
The Java Language Specification does not allow casting from a reference type to a primitive if it is not by unboxing conversion:
The compile-time legality of a casting conversion is as follows:
An expression of a primitive type may undergo casting conversion to another primitive type, by an identity conversion (if the types are the same), or by a widening primitive conversion, or by a narrowing primitive conversion, or by a widening and narrowing primitive conversion.
An expression of a primitive type may undergo casting conversion to a reference type without error, by boxing conversion.
An expression of a reference type may undergo casting conversion to a primitive type without error, by unboxing conversion.
An expression of a reference type may undergo casting conversion to another reference type if no compile-time error occurs given the rules in §5.5.1.
So the only primitive type to which a Long can be cast is long. If you declare the variable of type long (the primitive type), the cast will compile:
long l = (long) d.doubleValue();
int i = (int) l;