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.
Related
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}
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:
Can I pass an array as arguments to a method with variable arguments in Java?
(5 answers)
Closed 6 years ago.
I have a java program with a method that will consistently receive a different number of int values. What is the best way to send it the ints?
A construct called varargs (or an arbitrary number of arguments) helps you.
method(int... ints) { ... }
Then, varargs will be turned into int[] by the compiler.
OK, how to call those methods?
method(1, 2, 3);
method(new int[]{1, 2, 3});
About the question in the comments:
method(Arrays.stream(yourStringArray).mapToInt(Integer::parseInt).toArray());
You may convert String[] to int[] firstly, and then pass the result to the method.
Something like below you can do, making use of dot notation for method declaration with multiple arguments
public TestClass
{
public void process(int... variables)
{
// process your variables,
// Iterate through variables
}
}
Plus refer this, Java multiple arguments dot notation - Varargs
This question already has answers here:
Why is String[] args required in Java?
(6 answers)
Closed 9 years ago.
When I see Java programs, many leave the String args[] on even though the program doesn't use them. Why is this? Anything in particular?
String args[] is part of the method signature for main. If you don't have it you will get the exception below when you try and run the code.
Exception in thread "main" java.lang.NoSuchMethodError: main
If it is for a main(String[]) it is to fulfill the method signature & therefore vital.
Its required by the specification.
The method main must be declared public, static, and void. It must
specify a formal parameter (ยง8.4.1) whose declared type is array of
String. Therefore, either of the following declarations is acceptable:
public static void main(String[] args)
public static void main(String... args)
Specification
The code within the main may not directly use it, but it is still required. 'String args[]' is where any command line arguments are passed. Even if you pass in 0 arguments, there needs to be a way for that to be verified. It is also the required signature for main by the requirements of the JVM.
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.