问题
Today I found that the following code compiles and runs with no any warning:
public class Try_MultipleArguments2 {
public static void main(String[] args) {
myfunction();
myfunction(1, 2, 3);
}
public static void myfunction(int ... as) {
System.out.println("varags called");
}
public static void myfunction() {
System.out.println("noarg called");
}
}
I am remembering clear, that it was not so earlier.
Is this JVM change or my memory glitch???
How it distinguish between no-arg and varargs?
UPDATE
The following code also runs ok:
public class Try_MultipleArguments2 {
public static void main(String[] args) {
myfunction();
myfunction(1, 2, 3);
}
public static void myfunction(int ... as) {
System.out.println("varags called");
}
// public static void myfunction() {
// System.out.println("noarg called");
// }
}
回答1:
These are overloaded methods. The compiler knows which method the compiled main shoulld call from the method signature. See this specification:
When a method is invoked (§15.12), the number of actual arguments (and any explicit type arguments) and the compile-time types of the arguments are used, at compile time, to determine the signature of the method that will be invoked (§15.12.2).
Furthermore, the method chosen is the one that is most specific. See this. In this case, the no-arg method is more specific than the varargs version - again the number of parameters is checked to see which method to choose..
回答2:
Its function overloading at backend.
Your void myfunction(int ... as) is accepting multiple arguments while your void myfunction() has no argument. I don't see any glitch in this . Method Overloading
来源：https://stackoverflow.com/questions/31201196/when-varargs-started-to-not-conflict-with-no-arg