Pass object Java

Question

What is the difference between following code in Java? I couldn't clearly catch out how objects are passed in Java. Can anyone explain the code below.

package cc;

public class C {

    public static class Value {
        private String value;

        public void setValue(String str) {
            value=str;
        }

        public String getValue() {
            return value;
        }
    }



    public static void test(Value str) {
        str.setValue("test");
    }

    public static void test2(Value str) {
        str=new Value();
        str.setValue("test2");
    }
    public static void main(String[] args) {
        String m="main";
        Value v=new Value();
        v.setValue(m);
        System.out.println(v.getValue()); // prints main fine
        test(v);
        System.out.println(v.getValue()); // prints 'test' fine
        test2(v);
        System.out.println(v.getValue()); // prints 'test' again after assigning to test2 in function why?
    }
}

Java is *pass-by-value*. You passed the value of a reference to `v` from `main` into `test2` via the parameter `str`, which was lost when `str` was assigned to a new reference. — obataku, Aug 23 '12 at 06:55
possible duplicate of [Is Java "pass-by-reference"?](http://stackoverflow.com/questions/40480/is-java-pass-by-reference) — KV Prajapati, Aug 23 '12 at 07:02

score 3 · Accepted Answer · edited May 23 '17 at 12:04

3

In test2() a new instance is created and you are setting value in newer instance, and not in the Object whose reference is passed

public static void test2(Value str) {
        str=new Value();
        str.setValue("test2");
}

See Also

Is Java "pass-by-reference" or "pass-by-value"?

edited May 23 '17 at 12:04

Community

1
1

answered Aug 23 '12 at 06:55

jmj

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Its ok. But shouldn't the passed reference should point to new object created inside test2()? It is still pointing to old object even we changed inside the function. That is the thing I couldn't get it clearly? – Kusum Adhikari Aug 23 '12 at 07:01
No only value of reference are copied to new reference variable, So still v from main points to the object created in main, v (reference) its value is copied over to str and then new value is assigned to it when you create `new Value()` – jmj Aug 23 '12 at 07:02

score 1 · Answer 2 · answered Aug 23 '12 at 06:59

When you pass v into test2 what you are actually doing is passing a copy of the reference that points to v into the method - think of it as a new variable which has the same value as v.

When you assign a new Value object to str inside test2 you are assigning a new value to the copy of the reference - your original reference, which exists inside main is unaffected by this reassignment, and still refers to the original object.

When you execute your third println in main you are still using the original Value object, not the new one that was created inside test2, and so you see test and not test2 printed out.

score 0 · Answer 3 · answered Aug 23 '12 at 07:01

in java, object is passed by value. it copies the reference of v2 to str. Although you create a new object and assigned to str, str is local variable within test2 method scope. Only str inside test2 method is changed.

public static void test2(Value str) {
  str=new Value();
  str.setValue("test2");
  System.out.println(str.getValue());
}

Pass object Java

3 Answers3