Passing object and string literal as parameters

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Java

class Main {
static void m (int[] i) {
i[0] += 7;
}
public static void main (String[] args) {
int[] j = new int[1];
j[0] = 12;
m(j);
System.out.println(j[0]);
}
}

output
19

Here j and i points to the same address so the value gets changed for i too.

What I have tried:

Java

class Main {
static void m (int i) {
i += 7;
}
public static void main (String[] args) {
int j;
j= 12;
m(j);
System.out.println(j);
}
}

output
12

Java

class Main{
static void m(String s1) {
s1=new String("acting");

}
public static void main(String[] args) {
String s2 = new String("action");
m(s2);
System.out.println(s2);
}
}

output
action

here separate memory will be created for s1 and s2 so this output makes sense for me.

Java

class Main{
static void m(String s1) {
s1="acting";

}
public static void main(String[] args) {
String s2 ="action";
m(s2);
System.out.println(s2);
}
}

output
action

But here the string literals will be pointing to the same location and then too the value of s2 is not changed. Could anyone explain me clearly?

Posted 18-Jul-18 0:19am

@k5hu

Updated 18-Jul-18 0:40am

v2

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CPallini · Answer 1 · 2018-07-18T00:39:00

Solution 1

See here: Java Method Arguments[^].

Posted 18-Jul-18 0:39am

CPallini

Jochen Arndt · Answer 2 · 2018-07-18T00:41:00

With Java, all paramters are passed by value. That means a function gets a copy of the original reference when passing objects like strings. Because strings are Immutable Objects (The Java™ Tutorials > Essential Classes > Concurrency)[^], a new string is created and assigned to the copy of the reference while the original is left unchanged.

But when using an array, the copy of the reference is unchanged (still the same value as the original) so that array items can be changed.

See also the SO thread c - Passing a String by Reference in Java? - Stack Overflow[^] which also provides solutions to modify a string passed as function parameter.