Basic

In this topic, the most important thing is to make primitive and reference value clear.

1 2	int i = 1; String str = “Allen”

What are they like in the memory?

primitive
reference

1	str = "Java"

reference

Actually, they are all values. Primitive holds the value itself, while reference refers to address value.

Now let’s see some examples.

Example 1

static void test(int i, String str){
	i = 1;
	str += "1";
	System.out.println(str.hashCode());
}

public static void main(String[] args){
	int i = 11;
	String str = "Allen";
	System.out.println(str.hashCode());

	test(i, str);
	System.out.println(i);
	System.out.println(str);
}

Result:
63353322
1963953031
11
Allen

Explain
i and str are primitive type, so Java passes them as value copies into the method.
Observe the hashcode, string inside and outside test method 
are not the same.

str in main method
str in test method

Before str execute +=, it points to “Allen”. While doing +=, “Allen” cannot be changed in heap, so Java has to get a new memory block to hold “Allen1” in heap. Eventually, variable str’s hold a new address value 0x45 pointing to “Allen1”.
Additionally, when Java passes it into method, the variable str with reference address 0x34 is copied into the variable in the test method, which is called passing data by value in Java. Address or value itself are regarded as value in general, which is the most important to make sense in this topic.

Example 2

static void test(int i, StringBuilder str){
	i = 1;
	str.append("1");
	System.out.println(str.hashCode());
}

public static void main(String[] args){
	int i = 11;
	StringBuilder str = new StringBuilder("Allen");
	System.out.println(str.hashCode());

	test(i, str);
	System.out.println(i);
	System.out.println(str);
}
  
Result:
1625635731
1625635731
11
Allen1

Explain
Here str is object not primitive, so pass a copy into the method.
Observe the hashcode, same.