This question already has answers here:
Does Java support default parameter values?
(28 answers)
Closed 2 years ago.
I want to know if there is a way of makeing optional parameter in java.
for my experience in other programing languages there is an option to write function prototype like that:
int foo(int num = 0)
and if i call the func foo() the var num will be equal to 0.
I've been looking over the internet and its seem to be not exists in java.
Varargs should fulfill your request here.
Call method without passing any arguments or pass null parameter:-
foo(null) or foo().
Your method signature:-
void foo(String... name){//Business logic}
Related
This question already has answers here:
varargs and the '...' argument
(3 answers)
Closed 6 years ago.
I have gone through a note on var-args in java. I wondered what is the difference it make with the array parameter while calling a method.
public void doSomething(int[] a){
// some logic here
}
public void doSometing(int... a){
// some logic here
}
the above two methods were called by
int[] x={1,2,3,4,5};
doSomething(x);
is both of them are same or some difference exists?
and is it possible to overload these two methods?
The two method signatures are the same, and they do not allow overloading.
The only difference is that calling doSomething(1, 2, 3) with the vararg signature is allowed, while calling the same with the array signature results in an error.
This question already has answers here:
What do 3 dots next to a parameter type mean in Java?
(9 answers)
Closed 7 years ago.
public static void main (String... arg)
I have never encountered the ... part in the function definition. Could anyone give some insight into this notation?
The ... indicates that you are passing 0 or more arguments of the type and the method will access them as an array of objects of the type. You may pass them as an array or as a sequence of objects of the declared type.
I.e.:
In your main method use
String firstArg = arg[0];
to access the first argument.
Look at the documentation of varargs for more info.
This question already has answers here:
Definition of a method signature?
(7 answers)
Closed 7 years ago.
I read a book titled 'Object First with Java' and in page 7 the author mentioned that the method signature "provides information needed to invoke that method". And the the author gave the following example:
void moveHorizontal(int distance)
However, today when I was watching a video about C# on Pluralsight, the author said that "the return type of a method is not part of the method signature".
I'm confused now and would like know what is a method signature?
A method-signature is the part of the method based on which you overload / override it. It contains :
The method name.
The arguments passed to it.
It doesn't contain :
Scope / access modifier
return type.
Method signature is used in interfaces and in abstract classes, but we always define the method data type(return type). It will be something invaluable if the return type is not a part of the signature.
This question already has answers here:
Which overload will get selected for null in Java?
(3 answers)
Closed 8 years ago.
I want to know whether this is a valid overloading :
public class OverLoadingTest{
private void callFunction(Object object){
System.out.println("Printing Object");
}
private void callFunction(String string){
System.out.println("Printing String");
}
}
Further more, since someone asked me this question.
If I do like this,
OverLoadingTest test = new OverLoadingTest();
test.callFunction(null);
what will be printed ?
Of course my opinion is that it isn't valid overloading at all.
So no question of the second part.
Please tell me about this with some explanation.
The method with the least generic argument is called. So, in your case, it will be method accepting String
Note : If 2 classes are at the same level, then you will get an ambiguous call exception. For example if one method took String and another took Exception.
If more than one member method is both accessible and applicable to a method
invocation, it is necessary to choose one to provide the descriptor for
the run-time method dispatch.
The Java programming language uses the rule that the most specific method is chosen.
See more details in JSL 15.12.2.5
In your case, String method will be invoked, if argument is String or null and for other argument's types Object method will be invoked.
In your example, if you define one more method with argument type that is not String (e.g Integer), can't compile the source as it is ambiguous to be invoked between the methods with String and Integer as they are same level.
This question already has answers here:
varargs and the '...' argument
(3 answers)
Closed 9 years ago.
I have to implement an API for a homework assignment, and my instructor has used a notation I am unfamiliar with for one of the methods in the API (javadoc based).
public void method(String... strs);
What do the '...' mean? It later looks like I'll need to call this same method using a single string actual parameter, as well as multiple string actual parameters...
Java doesn't have optional arguments (to my knowledge), so I am a little confused here...
It's called varargs; http://docs.oracle.com/javase/6/docs/technotes/guides/language/varargs.html
It means you can pass an arbitrary number of arguments to the method (even zero).
In the method, the arguments will automatically be put in an array of the specified type, that you use to access the individual arguments.
Yes, that means you can take arbitrary no of Strings as an argument for this method.
For your method:
public void method(String... strs);
You can call it as:
method(str)
method(str1, str2)
method(str1,str2,str3)
Any no of arguments would work. In other words, it is a replacement for:
public void method(String[] str);
It's called an ellipsis and it means the method can take multple Strings as its argument.
See: The Java tutorial on passing arguments on Oracle's site.
See java optional parameters : as of Java 5, Java has support for variable numbers of arguments.