How to call a void method in the main?

Question

public void AmPm(int time) {
 if (time >= 12 && time < 12)
   System.out.println("AM");
 else if (time >= 12 && time < 24)
   System.out.println("PM");
 else
   System.out.println("invalid input");}

How can I call this method in the main method

Since this isn't static, you would first need to instantiate whatever class this method belongs to. — khelwood, Dec 01 '17 at 11:46
You already are using example of this. `println` is an instance `void` method which takes one argument. How did you call `println`? ... by using the instance `System.out` — Peter Lawrey, Dec 01 '17 at 11:56
BTW the simplest solution is to make this method `static` as it doesn't have to be an instance method. Again, your `main` is a `static` method so you know how to do this already. — Peter Lawrey, Dec 01 '17 at 11:57

Dawid Fieluba · Answer 1 · 2017-12-01T12:06:28.373

1

Main method is static, and from static methods you can call only static ones. What you can do is:

class A {
    public void amPm(int time) {
       if (time >= 0 && time < 12) //you have a typo there
           System.out.println("AM");
       else if (time >= 12 && time <24)
           System.out.println("PM");
       else
           System.out.println("invalid input");
    }
    //or as static method:
    public static void amPmInStaticWay(int time) { //... }

    public static void main(String[] args) {
        //...
        A a = new A();
        a.amPm(time);
        //or
        amPmInStaticWay(time);
        //or if you want to use static method from different class
        A.amPmInStaticWay(time);
    }
}

edited Dec 01 '17 at 12:06

answered Dec 01 '17 at 11:50

Dawid Fieluba

1,271
14
34

3

Yes, but as the method amPm only uses the parameter, you can rather mark it as static. – Maurice Perry Dec 01 '17 at 11:59
I can't change the code it is a homework XD – Atheer Dec 01 '17 at 12:17
Where you can't change it? – Dawid Fieluba Dec 01 '17 at 12:19
where? i didn't understand – Atheer Dec 01 '17 at 12:23
And i tried your code but it doesn't work :( – Atheer Dec 01 '17 at 12:24
1

Which changes do u mean? As long as it's _homework XD_ shouldn't it work properly (I assume that u mean this: `(time >= 0 && time < 12)` change. In your code there is no way that your method would print "AM" because u want input to be >=12 and < 12 in the same time. – Dawid Fieluba Dec 01 '17 at 12:26
As you cant to change the code, answer is simple- you cannot to call it in the main... Otherwise it has to been marked as static, or to be instaced and present in some object. – xxxvodnikxxx Dec 01 '17 at 12:32

score 1 · Answer 2 · answered Dec 01 '17 at 12:28

1

You need to create an instance of object of class A because this method isn't static. Then you can call that method on that reference:

public static void main(String[] args) {
    A a = new A();
    a.amPm(time); /* instead of typing "time" you need to pass int value that you want to be passed to the method as an argument */
}

answered Dec 01 '17 at 12:28

Przemysław Moskal

3,551
2
12
21

could you explain i couldn't understand:( – Atheer Dec 01 '17 at 14:17

score 0 · Answer 3 · answered Dec 01 '17 at 13:39

0

your method not static so you should create object from the class this method belongs to. The idea is : The static methods is at class level, so you can't call non static methods from static methods.

answered Dec 01 '17 at 13:39

Yasser Adnan

1
3

How to call a void method in the main?

3 Answers3