Explaination of method Overloading and Type Promotion in Java (Very importent for MNC tricky question)
Boolean---->Byte ---->Char ---->Short
---->int ---->long ---->float ---->double
Type promotion is an automatic type conversion from
a "lesser" base type to a "greater" one. Let us see
the diagram given below and understand it.
It must be clear from the above diagram that byte
can be promoted to short, int, long, float or double. The short data type can
be promoted to int, long, float or double. The char data type can be promoted
to int, long, float or double and so on.
One type is promoted to its greater type implicitly
it no matching data type is found.
a) Example of method overloading with Type
Promotion:
public class
TypePromotion {
|
void add(float a, float b) {
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System.out.println(a + b + " float");
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}
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void add(long a, long b) {
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System.out.println(a + b + " long");
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}
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public static void
main(String args[]) {
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TypePromotion obj = new
TypePromotion();
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obj.add(21, 21);
// integer value is passed
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}
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}
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Output: 42 long
In the above example, integer value is passed
through obj.add(21,21) which will be implicitly promoted to
its nearest bigger type i.e. long type.
b) Example of
method overloading with Type Promotion:
class
TypePromotion {
|
void add(float a, float b) {
|
System.out.println(a + b + " float");
|
}
|
void add(double a, double b) {
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System.out.println(a + b + " double");
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}
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public static void
main(String args[]) {
|
TypePromotion obj = new
TypePromotion();
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obj.add(21, 21);
// integer value is passed
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}
|
}
|
Output: 42.0
float
In the above example, an integer value is passed
through obj.add(21,21) which will be implicitly promoted to its
nearest bigger type i.e. float type. [Note: nearest bigger type
of int is long which is not matched in the above program,
therefore next bigger type is matched and is promoted
to float implicitly]
c) Example of
method overloading with Type Promotion
public class
OverloadingExm {
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void sum(int a, int b) {
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System.out.println("int arg method invoked");
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}
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void sum(long a, long b) {
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System.out.println("long arg method invoked");
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}
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public static void
main(String args[]) {
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OverloadingExm obj = new
OverloadingExm ();
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obj.sum(20, 20);// now int arg sum() method gets
invoked
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}
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}
|
Output:
int arg method invoked
d) Example of
method overloading with Type Promotion giving ambiguity
class
TypePromotion {
|
void add(int a, long b) {
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System.out.println(" 1st method invoked");
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}
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void add(long a, int b) {
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System.out.println(" 2nd method invoked");
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}
|
public static void
main(String args[]) {
|
TypePromotion obj = new
TypePromotion();
|
obj.add(21,
21);
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}
|
}
|
Compile time error will be generated
because one-type cannot be demoted to any type implicitly.
Output:
Exception in thread "main" java.lang.Error: Unresolved
compilation problem: The method add(int, long) is ambiguous for the type
TypePromotion at com.pcpl.ccl.TypePromotion.main(TypePromotion.java:12)
