What is the output for the below code ? public class Test extends Thread{ static String sName = "good"; public static void main(String argv[]){ Test t = new Test(); t.nameTest(sName); System.out.println(sName); } public void nameTest(String sName){ sName = sName + " idea "; start(); } public void run(){ for(int i=0;i < 4; i++){ sName = sName + " " + i; } } }

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  1. good

  2. good idea

  3. good idea good idea

  4. good 0 good 0 1

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Correct Option: A

AI Explanation

To answer this question, let's go through each option to understand why it is correct or incorrect:

Option A) good - This option is correct.

In the given code, the variable sName is initially assigned the value "good". The nameTest method is called with the sName variable as an argument. In the nameTest method, the sName parameter is concatenated with the string " idea" and then the start method of the Thread class is called. This starts a new thread of execution and invokes the run method defined in the Test class.

In the run method, a for loop is executed four times, where the value of i is appended to the sName string. However, the sName variable in the run method is not the same as the sName parameter in the nameTest method. Therefore, the changes made to sName in the run method do not affect the original sName variable.

After the thread is started, the main thread continues execution and prints the value of sName, which is still "good".

Option B) good idea - This option is incorrect. The output is not "good idea" because the changes made to sName in the run method do not affect the original sName variable.

Option C) good idea good idea - This option is incorrect. The output is not "good idea good idea" because the changes made to sName in the run method do not affect the original sName variable.

Option D) good 0 good 0 1 - This option is incorrect. The output is not "good 0 good 0 1" because the changes made to sName in the run method do not affect the original sName variable.

The correct answer is Option A) good. This option is correct because the value of sName remains unchanged after the execution of the code.

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