I have this following code and I need some clarification regarding the overloading concept.
class Overload{
static void print(Object o){
System.out.println("object");
}
static void print(String s){
System.out.println("string");
}
public static void main(String...args){
print("Hello");
}
}
when I execute this, the output is String. Eventhough object is the super class, why does it display string instead of object?
thanks in advance
The rule in this case is:
the more specific signature will apply
String is more specific than Object, that's why print(String) is used. For instance, try with print(null) in your main, you will see find out that such code will not even compile, since the compiler will have no way to distinguish whether null is a String or an Object
Check out Oracle Overloading Tutorial
There is called string print because you pass "Hello" which is String object.
String also extends regular Object but overloading calls also the most upper type.
Currently, this version of overloaded method executes:
static void print(String s){
System.out.println("string");
}
Since you are calling like this:
print("Hello");
Data conversions happen only if and only if correct version of match is not found. And exact match is found for compiler.
First JVM sees for the print(String), if not found it searches for print() with argument of superclass of String. If still not found, then super-class of super-class of String. And this algorithm continues until there is print(Object). Or reporting an error
Because when there is an exact match, during overloaded functions, JVM will not consider generalization. i.e it will look for the best fit to match the argument type. In this case "Hello" is an exact match to String type. Since String fits better in String than in Object. Hence it will not consider
static void print(Object o)
