Question about ex println.

Question

Marry christmas all. prepare for a stupid question.
After a look in API I see 10 PrintStream.println() all with diffrent arguments.
So for example I can do .
System.out.println("A simple String);
System.out.println(120)And the method with the rigth argument is called. Or I can do this.
System.out.println("A simple String" + 120);Can I do the same with my own overloaded methods. Some thing like.
public class ConvertNumber{
convert(int i){
convert(double d){
}ConvertBumber.convert(10 + 10.23);

Answer 1

MagnusT76 wrote:
jverd wrote:
However, you do know that this will simply call the (double d) version, right? It kind of looks like you might be thinking that your one call to convert will invoke both signatures. That won't happen, and it doesn't happen in PrintStream or any other class. That's not how overloading works.Yes that was just what i thougth. How does it work?How does what work?
When you call println(X)--or ANY METHOD(X)--the argument X is fully evaluated before the method is called.
so println(1 + 2.34) first evaluates 1 + 2.34. Since at least one of the operands to the plus operator is a double, the other is promoted to a double, and the double math 1.0 plus 2.34 is performed. Since the result is a double, its println(double) that gets called, with the arg 3.34.