将数组传递给方法 Java

Question

How can I pass an entire array to a method?

private void PassArray() {
    String[] arrayw = new String[4];
    //populate array
    PrintA(arrayw[]);
}

private void PrintA(String[] a) {
    //do whatever with array here
}

How do I do this correctly?

Answer 1

You do this:

private void PassArray() {
    String[] arrayw = new String[4]; //populate array
    PrintA(arrayw);
}

private void PrintA(String[] a) {
    //do whatever with array here
}

Just pass it as any other variable.
In Java, arrays are passed by reference.

Answer 2

Simply remove the brackets from your original code.

PrintA(arryw);

private void PassArray(){
    String[] arrayw = new String[4];
    //populate array
    PrintA(arrayw);
}
private void PrintA(String[] a){
    //do whatever with array here
}

That is all.

Answer 3

An array variable is simply a pointer, so you just pass it like so:

PrintA(arrayw);

Edit:

A little more elaboration. If what you want to do is create a COPY of an array, you'll have to pass the array into the method and then manually create a copy there (not sure if Java has something like Array.CopyOf() ). Otherwise, you'll be passing around a REFERENCE of the array, so if you change any values of the elements in it, it will be changed for other methods as well.

Answer 4

Important Points

• you have to use java.util package
• array can be passed by reference

In the method calling statement

• Don't use any object to pass an array
• only the array's name is used, don't use datatype or array brackets []

---

Sample Program

import java.util.*;

class atg {
  void a() {
    int b[]={1,2,3,4,5,6,7};
    c(b);
  }

  void c(int b[]) {
    int e=b.length;
    for(int f=0;f<e;f++) {
      System.out.print(b[f]+" ");//Single Space
    }
  }

  public static void main(String args[]) {
    atg ob=new atg();
    ob.a();
  }
}

---

Output Sample Program

1 2 3 4 5 6 7

Answer 5

There is an important point of arrays that is often not taught or missed in java classes. When arrays are passed to a function, then another pointer is created to the same array ( the same pointer is never passed ). You can manipulate the array using both the pointers, but once you assign the second pointer to a new array in the called method and return back by void to calling function, then the original pointer still remains unchanged.

You can directly run the code here : https://www.compilejava.net/

import java.util.Arrays;

public class HelloWorld
{
    public static void main(String[] args)
    {
        int Main_Array[] = {20,19,18,4,16,15,14,4,12,11,9};
        Demo1.Demo1(Main_Array);
        // THE POINTER Main_Array IS NOT PASSED TO Demo1
        // A DIFFERENT POINTER TO THE SAME LOCATION OF Main_Array IS PASSED TO Demo1

        System.out.println("Main_Array = "+Arrays.toString(Main_Array));
        // outputs : Main_Array = [20, 19, 18, 4, 16, 15, 14, 4, 12, 11, 9]
        // Since Main_Array points to the original location,
        // I cannot access the results of Demo1 , Demo2 when they are void.
        // I can use array clone method in Demo1 to get the required result,
        // but it would be faster if Demo1 returned the result to main
    }
}

public class Demo1
{
    public static void Demo1(int A[])
    {
        int B[] = new int[A.length];
        System.out.println("B = "+Arrays.toString(B)); // output : B = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
        Demo2.Demo2(A,B);
        System.out.println("B = "+Arrays.toString(B)); // output : B = [9999, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
        System.out.println("A = "+Arrays.toString(A)); // output : A = [20, 19, 18, 4, 16, 15, 14, 4, 12, 11, 9]

        A = B;
        // A was pointing to location of Main_Array, now it points to location of B
        // Main_Array pointer still keeps pointing to the original location in void main

        System.out.println("A = "+Arrays.toString(A)); // output : A = [9999, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
        // Hence to access this result from main, I have to return it to main
    }
}
public class Demo2
{
    public static void Demo2(int AAA[],int BBB[])
    {
        BBB[0] = 9999;
        // BBB points to the same location as B in Demo1, so whatever I do
        // with BBB, I am manipulating the location. Since B points to the
        // same location, I can access the results from B
    }
}

Answer 6

You got a syntax wrong. Just pass in array's name. BTW - it's good idea to read some common formatting stuff too, for example in Java methods should start with lowercase letter (it's not an error it's convention)

Answer 7

class test
{
    void passArr()
    {
        int arr1[]={1,2,3,4,5,6,7,8,9};
        printArr(arr1);
    }

    void printArr(int[] arr2)
    {
        for(int i=0;i<arr2.length;i++)
        {
            System.out.println(arr2[i]+"  ");
        }
    }

    public static void main(String[] args)
    {
        test ob=new test();
        ob.passArr();
    }
}

Answer 8

public static void main(String[] args) {

    int[] A=new int[size];
      //code for take input in array
      int[] C=sorting(A); //pass array via method
      //and then print array

    }
public static int[] sorting(int[] a) {
     //code for work with array 
     return a; //retuen array
}

Answer 9

In this way we can pass an array to a function, here this print function will print the contents of the array.

public class PassArrayToFunc {

    public static void print(char [] arr) {
        for(int i = 0 ; i<arr.length;i++) {
            System.out.println(arr[i]);
        }
    }

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        char [] array = scan.next().toCharArray();
        print(array);
        scan.close();
    }

}