Unreachable Vs Dead Code

[Raja Narasimha]

Hi, Why I am getting compile warning as given in the comments.

public class P199_Exception26_return_Ask {
    public static int main(String[] args) {
	if (true) {
	    System.out.println("-- 1 --");
	    return 1;
	}
	return 1; //Why this is Dead Code instead of Unreachable Code ??
    }
}

[Rikesh Desai]

Raja Narasimha wrote:Hi, Why I am getting compile warning as given in the comments.

public class P199_Exception26_return_Ask {
    public static int main(String[] args) {
	if (true) { //line 1
	    System.out.println("-- 1 --");
	    return 1; //line 2
	}
	return 1; //Why this is Dead Code instead of Unreachable Code ??
    }
}

Because of these two lines..

[Abimaran Kugathasan]

Check the difference between Dead code and Unreachable code.

[Raja Narasimha]

Hi, What I am trying to ask is, Line# 5 is guaranteed to execute. As soon as the control finishes executing Line 5, Control should come out of the method. My question is why it should not be "Unreachable code" instead of "Dead Code" ? Hi Abimaran, one of your links says "Unreachable code is sometimes also called dead code" which is confusing me more.

Abimaran Kugathasan wrote:Check the difference between Dead code and Unreachable code.

[Abimaran Kugathasan]

Both are different, Unreachable code : the source code of a program which can never be executed because there exists no control flow path to the code from the rest of the program, and Dead code : the source code of a program which is executed but whose result is never used in any other computation. So, for an example, the below code is an unreachable code,

import java.io.*;

public class TestUnreachableCode	{
	public static void main(String[] args) throws Exception	{
		int x = 3;
		throw new IOException();
		System.out.println(x);
	}
}

You can see a unreachable statement in the above code.

[Raja Narasimha]

Yes. In your program

throw new IOException();

is guanteed to execute, so you will get the error if you write any statement after that.

In my code, Lines 1-4 is guaranteed to execute and we are returning out of the method at Line 3. How can we return again at Line 5. It should not make any sense whether it is executed or not. Please let me know

if (true) { //line 1  
         System.out.println("-- 1 --");  
         return 1; //line 2  
     }
return 1;

I am assuming that the above 5 lines of code can be written as given below but this gives error

	System.out.println("-- 1 --");
	return 1;
	return 1;

[Abimaran Kugathasan]

Raja Narasimha wrote:Yes. In your program

throw new IOException();

is guanteed to execute, so you will get the error if you write any statement after that.

In my code, Lines 1-4 is guaranteed to execute and we are returning out of the method at Line 3. How can we return again at Line 5. It should not make any sense whether it is executed or not. Please let me know

if (true) { //line 1  
         System.out.println("-- 1 --");  
         return 1; //line 2  
     }
return 1;

I am assuming that the above 5 lines of code can be written as given below but this gives error

	System.out.println("-- 1 --");
	return 1;
	return 1;

What you have forgot is, the compiler don't know the logics, it knows only the syntax!