Method calling

[Ram Han]

class Inttest {
public static void main(String a[]){
Inttest t=new Inttest();
t.m1(0);
}
public void m1( int s){
System.out.println("Int ");
}
public void m1( short s){
System.out.println("Short ");
}
}


Here the output is printed as "Int "

Why not " Short " ?

[Philippe Saint-Just]

"Generally speaking, a series of digits with no decimal point is typed as an integer."

http://java.sun.com/docs/books/tutorial/java/nutsandbolts/datatypes.html

[vipul patel]

Read:
While matching the arguments of overloaded methods, compiler performs the closest match.

Literal 0, is considered as an integer. because by default all integral number literals are treated as integer only.

do some change as follow. it will print 'short'.


class Inttest
{
public static void main(String a[])
{
Inttest t=new Inttest();
t.m1((short)0);
}
public void m1( int s)
{
System.out.println("Int ");
}
public void m1( short s)
{
System.out.println("Short ");
}
}