Why does String passed by reference doesn't change its value?

Why does the piece of code bellow modifies only value in array t and doesn’t change value of the string s as well? I expected that String would also be changed because of its object properties.

class A {
    private int i = 0;
    private String s = "";
    private int[] t = new int[1];

    void m() {
        B.m(i, s, t);
    }

    @Override
    public String toString() {
        return "i=" + i + ", s=" + s + ", t[0]=" + t[0];
    }
}

class B{
   public static void m(int i, String s, int[] t){
      i += 1;
      s += "1";
      t[0] += 1;
   }  
}

public class Zad {

   public static void main( String [ ] args ){
      A a = new A() ;
      System.out.println(a);
      a.m();
      System.out.println(a);
   } 
}

>Solution :

This is what happening in your code: In Java, when you pass primitive types (like int) to a method, you pass the value itself. Any modification made to the parameter within the method do not affect the original value outside the method. This is known as "pass-by-value". This you may already know.

In definition of class b, i, s, and t are local variables within the method m(). It means, changes made to these variables do not affect the original values in the calling code.

When you call instance method a.m(); in your main method:

A a = new A();
System.out.println(a);  // Output: i=0, s=, t[0]=0
a.m();
System.out.println(a);  // Output: i=0, s=, t[0]=1

You’ll see that the value of t[0] has changed because arrays in Java are objects, and modifications to their elements are reflected outside the method. However, the values of i and s remain unchanged because they are primitive types or immutable objects (like String), and any changes made to them within the method are local and don’t affect the original values.