What order are fields initialized in?

Question

I read this question while studying model questions for exams.

class Was { private int a=show(); private int b=5; private int show() { return b; } public static void main(String args[]) { System.out.println((new Was()).a); } }

I compiled this,it prints as 0. Why it is not print as '5' ?

Because the initializers are evaluated in the order they appear. — Hot Licks
– Hot Licks, Commented Dec 22, 2013 at 19:25

Sotirios Delimanolis · Accepted Answer · 2013-12-22 19:25:15Z

4

Field initialization happens when a constructor is invoked and in order of declaration. This field

private int a=show();

is initialized before

private int b=5;

It is initialized with the value returned by show(). At that time b has yet to be initialized to 5. Its default value is 0. So show() returns 0. That value is assigned to a.

answered Dec 22, 2013 at 19:25

Sotirios Delimanolis

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3 Comments

anavaras lamurep Over a year ago

Thank you @Sotirios Delimanolis.But,at the time of calling show(), b is unknown field,how it can be returned?

Sotirios Delimanolis Over a year ago

@anavaraslamurep The field is known, it has just been given the default value for int primitives, which is 0.

Patricia Shanahan Over a year ago

Java object fields go through two steps. Before any constructor or initializer execution, each is initialized to a default value that depends only on its type, 0 for numeric fields, null for references, false for booleans. That value can be changed later by an initializer or by constructor code.

Suresh Atta · Accepted Answer · 2013-12-22 19:25:35Z

private int a=show();

By the time you calling show() method the value of b not yet set and hence the default value of integer is 0 results.

To see desired output

private int b=5; private int a=show();

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