Why can't we initialize local variable before declaring it unlike static variable?

Question

Why can't i be initialized before declaring it in code 1 but it can be in code 2 :

    class Code1 {

        void methodOfCode1() {
        //    System.out.println(i); can't use a local field before it is defined
        //    i = 10;                can't initialize a local field before it is defined
              int i;
        }
    }



class Code2{
    static
    {
        i = 10;   //A field can be initialized before it is defined.
    }


    static int i;
}

Because the compiler won't recognize the variable when you're initializing it. Take a look at [this](http://java.about.com/od/understandingdatatypes/a/declaringvars.htm) for details. — DCruz22, Jun 01 '16 at 13:03
Java is a strongly typed language, meaning it has to know what kind of values each variable is going to hold before assigning them a value. — Orin, Jun 01 '16 at 13:04

score 3 · Accepted Answer · edited May 23 '17 at 12:22

3

The answer is order of execution.

Code executes sequentially, you must define what something is before you use it. Therefore in a method you must declare a variable before you use it.

In Java the compiler initializes your class in multiple passes. It first runs through and declares all of the static members, then it sets the default values for all static members. This all happens before any methods (static or not) are executed.

edited May 23 '17 at 12:22

Community

1
1

answered Jun 01 '16 at 13:09

cyroxis

3,661
22
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Great explanation!! Thank you. – sms sam Jun 01 '16 at 13:19

Vishal Gajera · Answer 2 · 2016-06-01T13:23:03.480

0

In Java, a variable's type must be declared before it can be used.

For example:

System.out.println(i); // for this statement i is unknown/undefined...
i = 10;  // same for above...

EDITED :

First inside class Code1,

you haven't declare variable before use it, which is against of Java Standard API so obviously compiler will complain.

while into class Code2 case,

you have declare 'static' both instance as well as block.

so it does not required any instance to load it. (review static concept in depth will let you to the correct direction)

edited Jun 01 '16 at 13:23

answered Jun 01 '16 at 13:04

Vishal Gajera

4,137
5
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Then why i can be initialize in below case?Please elaborate a little( i am new to java) public class A { { i = 20; //but, can be initialized } int i; } – sms sam Jun 01 '16 at 13:06
@smssam what you commented is different than what you have asked into this Question... please clear your question first, we will provide you clear answer. – Vishal Gajera Jun 01 '16 at 13:10
I have edited the question. Why can't i initialize i before declaring it in code 1 but the same can be done in code 2. – sms sam Jun 01 '16 at 13:17
@smssam see my EDITED section into my existence Answer. – Vishal Gajera Jun 01 '16 at 13:23
Thank you @vishal, that was indeed helpful. – sms sam Jun 01 '16 at 13:37

score 0 · Answer 3 · edited Jun 01 '16 at 13:15

Java is strongly-typed, you can't use something that you don't know what it is. For instance, a method is different if it receives a String or if it receives an int, so the compiler must be able to know which is the proper method to be invoked

For instance, this is valid:

public void dealWithIt(int val){
   val++
   System.out.println(val);
}

public void dealWithIt(String val){
    System.out.println(val);
}

The compiler can't assign a value to a variable before know what type of variable it is.

score 0 · Answer 4 · answered Jun 01 '16 at 13:06

It's the scope of the variable & the lack of type prior to its use: you have yet to define i. You must define a variable before it can be used:

int i = 0; 
System.out.println(i); // works
System.out.println(j); // will not work
byte j;

Here's the Java Language Specification giving you the detail.

score 0 · Answer 5 · answered Jun 01 '16 at 13:07

0

During initialise a variable, some amount of memory is allocated based on datatype.

Unless you allocate memory, you can't initialize value.

answered Jun 01 '16 at 13:07

PriVictor

49
5

score 0 · Answer 6 · answered Jun 01 '16 at 13:11

Because compiler must know where this variable is stored and how many bytes it have to use. If I tell computer var=97 before initializing, it cannot know where this "var" variable is stored. Also if "var" is char , computer needs one byte to store in RAM. However it also can be an integer which is 4 bytes(depends on different systems). Therefore, there would be an ambiguous situation for the system.Compiler does not allow that , reasonably.

Why can't we initialize local variable before declaring it unlike static variable?

6 Answers6