Why can't I call a method with an array variable?

Question

I am trying to print out a simple code but I do not seem to be able to pass the array variable of the method. Sorry if it is something obvious, I am only starting with Java. I am getting the "The method asd in the type MyClass is not applicable for the arguments (int,int,int,int)

public int asd(int[] nums) {
      int count = 0;
      // Note: iterate to length-1, so can use i+1 in the loop
      for (int i=0; i < (nums.length-1); i++) {
        if (nums[i] == 6) {
          if (nums[i+1] == 6 || nums[i+1] == 7) {
            count++;
          }
        }
      }
      return count;
    }

public static void main(String[] args) {
    // TODO Auto-generated method stub

    System.out.println(asd(1,22,3,4,2,2,2));

}

I'm not sure why this was never mentioned, but you can declare a variadic argument function by making it look like this: public int asd(int... nums) { The ... can only be on the last argument in a method declaration as it collects all arguments after that one into an array. — Powerlord
– Powerlord, Commented Oct 8, 2024 at 3:04

prasad_ · Accepted Answer · 2024-10-08 05:17:22Z

5

The method asd is expecting a single parameter of type int []. You are attempting to pass each integer as a separate parameter (resulting in seven parameters , 1,22,3,4,2,2,2).

To correct the compiler-error you can enclose the parameter values in {} and prefix it with new int[] to create an Array literal:

 System.out.println(asd(new int[] {1,22,3,4,2,2,2}));

edited Oct 8, 2024 at 5:17

prasad_

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answered Jun 19, 2019 at 0:49

user10316640

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1 Comment

subhashis Over a year ago

I will give you the answar, wait for me. :)

subhashis · Accepted Answer · 2024-10-11 01:49:37Z

-1

you can try this

public static void main(String[] args) {
    int []arr={1,22,3,4,2,2,2}; //this is array literal
    System.out.println(asd(arr));
}

The above program will give compilation error attaching screenshot below

edited Oct 11, 2024 at 1:49

answered Oct 8, 2024 at 2:51

subhashis

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15 Comments

user85421 Over a year ago

not really different then the previous (5 year old) answer, is it? But for much less explanation (none!)

subhashis Over a year ago

It will give compilation error, in any IDE. And why I gave a new answar because where is code reusability?

user85421 Over a year ago

1) which compilation error? It is valid Java (if we ignore that asd is not static); 2) what do you mean with "* code reusability*"? How is that relevant, does that apply here?

subhashis Over a year ago

I have added the screenshot see the error, and for code reusablity we have to discuss and I dont have time for it figure it out. I just copy and paste the code.

Stephen C Over a year ago

(And if you don't have time to post a proper answer ... on an old question ... then the solution is don't post an answer at all.)

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