how many arguments should at least passed to a var-args in a method?

[prakash Venugopal]

class VarargExa
{
public static void main(String...args)
{
go();
}
static void go(short...args)
{
System.out.println("Inside of go");
}
}


The out put is: Inside of go

[saloni jhanwar]

0 to many.

[gurpeet singh]

prakash Venugopal wrote:class VarargExa
{
public static void main(String...args)
{
go();
}
static void go(short...args)
{
System.out.println("Inside of go");
}
}


The out put is: Inside of go

Would like to add to what saloni said. when you pass arguments to method which accepts var-args they are passed in the form of an array of the type of the var-arg declared. in above case if you invoke go like for e.g
go(2,3,4) the arguments will be passed as short array. you can say that the var-args argument is nothing but array of shorts , because of which you can write main method signature as :

main(String[] args) or main(String... args)

which means if you try to overload the methods as go(short...args) and go(short[] s) it will give you error that the method name already exists.

[saloni jhanwar]

gurpeet singh wrote:

prakash Venugopal wrote:class VarargExa
{
public static void main(String...args)
{
go();
}
static void go(short...args)
{
System.out.println("Inside of go");
}
}


The out put is: Inside of go

Would like to add to what saloni said. when you pass arguments to method which accepts var-args they are passed in the form of an array of the type of the var-arg declared. in above case if you invoke go like for e.g
go(2,3,4) the arguments will be passed as short array. you can say that the var-args argument is nothing but array of shorts , because of which you can write main method signature as :

main(String[] args) or main(String... args)

which means if you try to overload the methods as go(short...args) and go(short[] s) it will give you error that the method name already exists.

Nice.