Here is what I know about overload resolution in java:
The process of compiler trying to resolve the method call from given overloaded method
In Java, resolving methods in case of method overloading is done with the following precedence:
1. Widening
2. Auto-boxing
3. Var-args
The java compiler thinks that widening a primitive parameter is more desirable than performing an auto-boxing operation.
In other words, as auto-boxing was introduced in Java 5, the compiler chooses the older style(widening) before it chooses the newer style(auto-boxing), keeping existing code more robust. Same is with var-args.
In your 1st code snippet, widening of reference variable occurs i.e,
IntegertoObjectrather than un-boxing i.e,Integertoint. And in your 2nd snippet, widening cannot happen fromIntegertoStringso unboxing happens.
Consider the below program which proves all the above statements:
class MethodOverloading {
static void go(Long x) {
System.out.print("Long ");
}
static void go(double x) {
System.out.print("double ");
}
static void go(Double x) {
System.out.print("Double ");
}
static void go(int x, int y) {
System.out.print("int,int ");
}
static void go(byte... x) {
System.out.print("byte... ");
}
static void go(Long x, Long y) {
System.out.print("Long,Long ");
}
static void go(long... x) {
System.out.print("long... ");
}
public static void main(String[] args) {
byte b = 5;
short s = 5;
long l = 5;
float f = 5.0f;
// widening beats autoboxing
go(b);
go(s);
go(l);
go(f);
// widening beats var-args
go(b, b);
// auto-boxing beats var-args
go(l, l);
}
}
The output is:
double double double double int,int Long,Long
Just for reference, here is my blog on method overloading in Java.
P.S: My answer is a modified version of an example given in SCJP.