Primitive Data Types

What is the result of executing the following code:
class Test {
public static void main(String[] args) {
Test t = new Test();
t.test(1.0, 2L, 3);
}
void test(double a, double b, short c) {
System.out.println("1");
}
void test(float a, byte b, byte c) {
System.out.println("2");
}
void test(double a, double b, double c) {
System.out.println("3");
}
void test(int a, long b, int c) {
System.out.println("4");
}
void test(long a, long b, long c) {
System.out.println("5");
}
}
Select the one right answer.
1
2
3
4
5
The answer is given as 3. I thought both 1 and 3 seem applicable. Cannot the constant 3 fit into a short? Can someone please explain?

Parimala!
Always constants match to integers.
that's why u get the answer as 3.
try this code
now you see the output as 1.
unless until u pass a short variable,
method test(double a, double b, short c) will be not be invoked
public static void main(String[] args)
{
MyTest t = new MyTest();
short s = 3;
long l = 4;
//t.test(1.0, 2L, 3);
t.test(2.0,l,s);
}
cheers
Siva

I am still not very clear about this.
Aren't the statements
short s = 3;
or
byte b = 3;
valid? Passing primitive type variables to a method is also similar. Isn't it?

Parimala read the content in my post again
Always constants without any suffixes like L, f, d are treated as integers only
test(2.0,3L,5);
is different from
test(f,l,s);
where
float f = 2.0f;
long l = 6;
short s = 9;
here f,l,s are variables.
incase of the first method, the arguments passed are not variables they are constant values. i.e. primitive literals.
Still not clear?
Siva

Thanks Siva. I understand your point.
But isn't there a rule that in an assignment statement like
final int i=3;
short s=i; // valid though no cast is present
is valid because although we assign a type int to type short, i is a final( and so constant) value?
Similarly, when we pass 3(which though integer is a constant) as parameter to a short type (s in the method), it should be valid.

Hi Parimala,
For assignments the compiler uses a special case so that the following can happen without casting:
short s = 100;
byte b = 56;
char c = 43;
The special case doesn't exist for parameter mapping. The compiler figures you better know what you want before you start passing things around. The compiler will only automatically perform widening while passing but never automatically perform narrowing conversions. Those are left to the programmer!
Regards,
Manfred.

Why is 2L and 3 reconized as a double?
Shouldn't there be a method that takes a double, long, int
I'm sorry for asking for another explanation but I can't get my head around what your saying.

Jonathan !
It would have been appropriate if there was a method that takes double long int.
Since its not there, as Manfred has said, the compiler does only widening conversion while passing parameters. so it cannot map 3 (int) to short.
Try taking out the double double double method. It would not even compile.
Hope this helps you.
[This message has been edited by Deepak Israel (edited April 03, 2001).]
[This message has been edited by Deepak Israel (edited April 03, 2001).]

Thans Manfred.