Doğa Can Yanıkoğlu Doğa Can Yanıkoğlu - 1 year ago 67
Java Question

The connection between 'System.out.println()' and 'toString()' in Java

What is the connection between

in Java? e.g:

public class A {
String x = "abc";

public String toString() {
return x;

public class ADemo {
public static void main(String[] args) {
A obj = new A();

If main class runs, it gives an output as
. When I remove the code which overrides
, it gives an output as
. So, can anyone explain what is the working principle of
when i use the parameter as an object? Is it fully connected with

Answer Source

System.out is a PrintStream. Printstream defines several versions of the println() function to handle numbers, strings, and so on. When you call PrintStream.println() with an arbitrary object as a parameter, you get the version of the function that acts on an Object. This version of the function

...calls at first String.valueOf(x) to get the printed object's string value...

Looking at String.valueOf(Object), we see that it returns

if the argument is null, then a string equal to "null"; otherwise, the value of obj.toString() is returned.

So, long story short, System.out.println(someObject) calls that object's toString() function to convert the object to a string representation.

If your object defines its own toString() function, then that is what will be called. If you don't provide such a function, then your object will inherit toString() from one of its parent classes. In the worst case, it will inherit Object.toString(). That version of toString() is defined to return

a string consisting of the name of the class of which the object is an instance, the at-sign character `@', and the unsigned hexadecimal representation of the hash code of the object.

Or, in other words:

getClass().getName() + '@' + Integer.toHexString(hashCode())

So, when you call System.out.println() on an object that doesn't define its own version of toString(), you might get the Object version which looks like "classname@someHexNumber".

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