Why this Output?

[Joe Harry]

Hi Ranchers,

Consider 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");
}
}

Why it prints 3 instead of 2?

Regards,
Jothi Shankar Kumar. S

[Keith Lynn]

Originally posted by Jothi Shankar Kumar Sankararaj:
Hi Ranchers,

Consider 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");
}
}

Why it prints 3 instead of 2?

Regards,
Jothi Shankar Kumar. S

There is only one method which matches the call.

[Anand Loni]

Hi,

2 is not printed because third parameter is integer and it can not converted to byte without explicit cast.

Regards,
Anand

[Joe Harry]

Hi Anand,

your quote:

Hi,

2 is not printed because third parameter is integer and it can not converted to byte without explicit cast.

Regards,
Anand

The third parameter's value is 3 which very well fits into a byte??So why not it take the 2nd method and print 2 at the output?

Regards,
Jothi Shankar Kumar. S

[Keith Lynn]

Method invocation conversion does not include narrowing primitive conversions.

3 is an int. It cannot be converted to a byte when it is sent as a parameter to a method that expects a byte.

[Joe Harry]

Hi Keith,

So when it comes to method invocation, any number of type 3.0, 55.0 is treated as double and numbers like 45, 888888 are treated as int. am I right?

Regards,
Jothi Shankar Kumar. S

[Sharmila Rishi]

So when it comes to method invocation, any number of type 3.0, 55.0 is treated as double and numbers like 45, 888888 are treated as int. am I right?

Yes, you are right.

Any floating point point value without a 'f' is implicitly a double and any numeric value is implicitly an int.


-Sharmila Rishi.

[Joe Harry]

Hi Sharmila,

Thanks for the comments.

Regards,
Jothi Shankar Kumar. S