(java) Determine the output for the following program: Determine the output for

Question

(java) Determine the output for the following program:

Determine the output for the following program: Private static void f (string a) {Try {Int x = 7/a. length(); System. out. PrintIn ("G");} catch (ArithmeticException e) {Sysyetm. out. PrintIn("F");} finally {System. Out. Printin("E");}} Public static void main (string[] args) {try {f("HELLO"); f(" "); f(null); system. out. printIn("D");} catch (Null pointerException e) {System. out. printIn("C");} catch (ArithmeticException e) {System. out. printin("B");} system. out. printIn("A");}

Explanation / Answer

The output of this program is as follows:

G
E
F
E
E
C
A

Description:

Execution of this program will start from main().

Let's first check what is the task of function f().

It takes a string, and divides 7 with its length.
This statement is written in try block.

If this division is successful, it will write "G".

If the string length was 0, it will be like 7 / 0 which will throw an exception of divide by zero which is an ArithmeticException.

If this exception is caught, It is printing "F". So, "G" will not be written this time.

And inside finally block, "E" is printed.

So, let's check the execution step by step.

The first line of main() is f("Hello") inside try block.

"Hello" string is passed to f() and the length of it is 5 which is not zero. So, this is the normal execution there and so "G" is printed and no exception is generated.

After that, finally block is going to run for every time, whether there was an exception or not. So, it prints "E".

The next line is f("") which means string with length 0 is passed to function.

So, here the length of string 'a' is 0. So, 7 / 0 throws an ArithmeticException. It is caught in the catch block and so it is printing "F".

After that, as always, "E" is printed at last in the function.

As this exception is already caught, it will not be thrown in the main() method. So, the ArithmeticException is not caught in main(). So, main() method catch block will not be executed for a single time.

The next line is f(null) so null is passed to function f() as 'a'.

And next thing done is finding a.length() but a = null here so this is throwing NullPointerException.

This statement was written inside try block so this exception is supposed to be caught, but there is no appropriate catch block for NullPointerException so this function executes its finally block which prints "E" and this exception is still thrown to main().

f(null) was written inside try block and it throws a NullPointerException right, so inside main(), if there is a catch statement, it will be handled there.

So, yes, the main() has a catch statement for NullPointerException and it is for same try in which f(null) is written,

So, that catch block is processed. And inside it "C" is printed.

And now that try catch is also completed. Here, in main(), there is 1 more statement that prints "A" so "A" is printed.

So, this is the sequence of execution.

Do comment if there is any query. Thank you. :)

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.