* Understanding do-while Execution
* A do-while loopexecutes at least oncebefore checking the condition.
* ++counter < 3 increments counterbeforeevaluating the condition.
* Step-by-Step Execution
* Iteration 1:counter = 0, print "0", then ++counter becomes 1, condition 1 < 3 istrue.
* Iteration 2:counter = 1, print "1", then ++counter becomes 2, condition 2 < 3 istrue.
* Iteration 3:counter = 2, print "2", then ++counter becomes 3, condition 3 < 3 isfalse, so loop exits.
* Final Output
0 1 2
Thus, the correct answer is:0 1 2
References:
* Java SE 21 - Control Flow Statements
* Java SE 21 - do-while Loop

In Java, a lambda expression can have two forms for its body:
* Single Expression:A concise form where the body consists of a single expression. The result of this expression is implicitly returned.
Example:
java
(a, b) -> a + b
In this example, (a, b) are the parameters, and a + b is the single expression that adds them together.
* Statement Block:A more detailed form where the body consists of a block of statements enclosed in braces {}. Within this block, you can have multiple statements, and if a return value is expected, you must explicitly use the return statement.
Example:
java
(a, b) -> {
int sum = a + b;
System.out.println("Sum is: " + sum);
return sum;
}
In this example, the lambda body is a statement block that performs multiple actions: it calculates the sum, prints it, and then returns the sum.
Given the options:
* A. Two statements:While a lambda body can contain multiple statements, they must be enclosed within a statement block {}. Simply having two statements without braces is not valid syntax for a lambda expression.
* B. An expression and a statement:Similar to option A, if a lambda body contains more than one element (be it expressions or statements), they need to be enclosed in a statement block.
* C. A statement block:This is correct. A lambda expression can have a body that is a statement block, allowing multiple statements enclosed in braces.
* D. None of the above:This is incorrect since option C is valid.
* E. Two expressions:As with options A and B, multiple expressions must be enclosed in a statement block to form a valid lambda body.
Therefore, the correct answer is C: A statement block.

In this code, the try block throws an IOException. The catch block catches this exception and throws a new RuntimeException. Regardless of exceptions thrown in the try or catch blocks, the finally block is always executed. In this case, the finally block throws an ArithmeticException.
When an exception is thrown in a finally block, it overrides any previous exceptions that were thrown in the try or catch blocks. Therefore, the ArithmeticException thrown in the finally block is the exception that propagates out of the method. As a result, the program terminates with an ArithmeticException.

1. Why does the compilation fail?
* The interface Special contains bar as int bar = 1;.
* In Java, all interface fields are implicitly public, static, and final.
* This means that bar is a constant (final variable).
* The method add() contains bar--, which attempts to modify bar.
* Since bar is final, it cannot be modified, causing acompilation error.
2. Correcting the Code
To make the code compile, bar must not be final. One way to fix this:
java
class SpecialImpl implements Special {
int bar = 1;
}
Or modify the add() method:
java
int add() {
return --foo + bar; // No modification of bar
}
Thus, the correct answer is:Compilation fails.
References:
* Java SE 21 - Interfaces
* Java SE 21 - Final Variables

In this code, an ExecutorService is created with a fixed thread pool of size 2 using Executors.
newFixedThreadPool(2). A Runnable task is defined to print "Task is complete" to the console.
The sequence of operations is as follows:
* service.submit(task);
This submits the task to the executor service for execution. Since the thread pool has a size of 2 and no other tasks are running, this task will be executed promptly, printing "Task is complete" to the console.
* service.shutdown();
This initiates an orderly shutdown of the executor service. In this state, the service stops accepting new tasks