I am new to java generics my question is :
public static < E > void printArray( E[] inputArray )
In the above statement when the retu
The <E>
here has nothing to do with the return type; it means that this is a generic function which can take various types of arrays. To be easier to understand, the code could be something like this:
public class GenericMethodTest
{
// generic method printArray
public static < E > void printArray( E[] inputArray )
{
// Display array elements
for ( E element : inputArray ){
System.out.printf( "%s ", element );
}
System.out.println();
}
public static void main( String args[] )
{
// Create arrays of Integer, Double and Character
Integer[] intArray = { 1, 2, 3, 4, 5 };
Double[] doubleArray = { 1.1, 2.2, 3.3, 4.4 };
Character[] charArray = { 'H', 'E', 'L', 'L', 'O' };
System.out.println( "Array integerArray contains:" );
GenericMethodTest.<Integer>printArray( intArray ); // pass an Integer array
System.out.println( "\nArray doubleArray contains:" );
GenericMethodTest.<Double>printArray( doubleArray ); // pass a Double array
System.out.println( "\nArray characterArray contains:" );
GenericMethodTest.<Character>printArray( charArray ); // pass a Character array
}
}
So what the <E>
does it tell the function "There is a generic type E
and you accept arrays of E
, so if for example I call you with the generic type Integer
you accept Integer[]
."
It does not change the return type though. Just like public static
doesn't change the return type.
By <E>
you define a generic variable type, if you will write
public static void printArray( E[] inputArray )
Java will try to find a class with name E
, because you don't define E
as a generic.
The return value in both cases are null, this <E>
generic definition part in not a part of return value, it's a separate block of function signature.
In public static < E > void printArray( E[] inputArray )
,
the E
is just used to signify a Generic parameter. The return type of the method is still void
.