xap/SelfCode · Datasets at Hugging Face

Code Line stringclasses 57 values	User Explanation stringclasses 692 values	Line-Explanation in PCEX stringclasses 131 values	Annotation Score float64 1 5 ⌀
Scanner scan = new Scanner(System.in);	creates a new Scanner instance which points to the input stream passed as argument.	To read the input value from the user, we need to define a Scanner object.	4
Scanner scan = new Scanner(System.in);	creates a new Scanner instance which points to the input stream passed as argument.	We need to read and process the value that the user enters.	2
Scanner scan = new Scanner(System.in);	In this case the steam is Standard input stream	To read the input value from the user, we need to define a Scanner object.	2
Scanner scan = new Scanner(System.in);	In this case the steam is Standard input stream	We need to read and process the value that the user enters.	2
System.out.println("Enter an integer for seconds: ");	prints the argument ("Enter an integer for second: ") passed to the standard console and a newline	We prompt the user to enter the seconds.	4
int seconds = scan.nextInt();	scaner object reads in a string of digits (characters) and converts them into an int type, and put them in variable entitled "seconds"	We need to read the seconds that the user enters and store it in a variable.	4
int seconds = scan.nextInt();	scaner object reads in a string of digits (characters) and converts them into an int type, and put them in variable entitled "seconds"	We read the seconds by calling the nextInt() method because this input is an integer.	4
scan.close();	closes scan the input that has been opened	We close the scanner as we do not want to process any input from the user in the rest of the program.	3
int minutes = seconds / 60;	creates varible entitled "minutes" which contain calculation of variable "second" divided by 60	To obtain the minutes in seconds, we divide the seconds by 60 because there are 60 seconds in a minute.	4
int minutes = seconds / 60;	creates varible entitled "minutes" which contain calculation of variable "second" divided by 60	Note that since both operands of division operator are integer, the fractional part of the result is truncated, if there is any.	2
int remainingSeconds = seconds % 60;	creates varible entitled "remainingSeconds" which contain calculation of variable "second" modulus by 60	This is because there are 60 seconds in a minute.	2
int remainingSeconds = seconds % 60;	creates varible entitled "remainingSeconds" which contain calculation of variable "second" modulus by 60	Note that the % operator returns the remainder of the division.	2
int remainingSeconds = seconds % 60;	creates varible entitled "remainingSeconds" which contain calculation of variable "second" modulus by 60	To obtain the remaining seconds after taking away the minutes, we have to take the remainder of the seconds divided by 60.	3
System.out.println(seconds + " seconds is " + minutes + " minutes and " + remainingSeconds + " seconds.");	prints in the console line the character passed inside the argument	This statement prints to the default standard output stream the minutes and remaining seconds from the input amount of time in seconds.	2
System.out.println(seconds + " seconds is " + minutes + " minutes and " + remainingSeconds + " seconds.");	prints in the console line the character passed inside the argument	The printed text is followed by the end-of-line character at the end.	1
int num = 15;	create variable integer entitled "num" with value 5	We define variable num to store the number that we want to find its smallest divisor.	3
int num = 15;	create variable integer entitled "num" with value 5	We could initialize it to any positive integer greater than 1.	2
int num = 15;	create variable integer entitled "num" with value 5	In this program, we initialize variable num to 15.	2
int divisor = 2;	creates variable integer entitled "division" with initial value 2	We initialize variable divisor by 2 because we want to find the smallest divisor except 1.	3
int divisor = 2;	creates variable integer entitled "division" with initial value 2	We define variable divisor to store the smallest divisor of the number.	2
int num = 15;	creates variable integer entitled "num" with initial value 5	We define variable num to store the number that we want to find its smallest divisor.	2
int num = 15;	creates variable integer entitled "num" with initial value 5	We could initialize it to any positive integer greater than 1.	2
int num = 15;	creates variable integer entitled "num" with initial value 5	In this program, we initialize variable num to 15.	2
int divisor = 2;	creates variable integer entitled "divisor" with initial value 2	We initialize variable divisor by 2 because we want to find the smallest divisor except 1.	3
int divisor = 2;	creates variable integer entitled "divisor" with initial value 2	We define variable divisor to store the smallest divisor of the number.	3
while (num % divisor != 0) {	creates looping "while" with condition while variable num modulus by the value of divisor is not null then execute the below code	Since we don't know ahead of time how many times the loop will be repeated, we need to use a while loop.	3
while (num % divisor != 0) {	creates looping "while" with condition while variable num modulus by the value of divisor is not null then execute the below code	We need to increment the divisor repeatedly as long as the divisor is not a factor of the number.	1
while (num % divisor != 0) {	creates looping "while" with condition while variable num modulus by the value of divisor is not null then execute the below code	Therefore, we need to use a loop structure.	4
while (num % divisor != 0) {	creates looping "while" with condition while variable num modulus by the value of divisor is not null then execute the below code	The condition in the while loop tests whether the body of the loop should be repeated, so it should test whether the divisor is not a factor of the number.	2
while (num % divisor != 0) {	creates looping "while" with condition while variable num modulus by the value of divisor is not null then execute the below code	We could check whether the divisor is not a factor of the number by computing the remainder of the division of the number by the divisor.	2
divisor += 1;	add divisor variable by + 1	When the divisor is not a factor of the number, we increment the variable divisor by 1.	3
System.out.println("The smallest divisor of " + num + " is " + divisor);	prints the character to the console with the argument inside it.	This statement prints to the default standard output stream the smallest divisor of the number.	2
Scanner scan = new Scanner(System.in);	defines a new scanner, named "scan" and inputs System.in	To read the input value from the user, we need to define a Scanner object.	3
Scanner scan = new Scanner(System.in);	defines a new scanner, named "scan" and inputs System.in	We need to read and process the integer that the user enters.	2
System.out.println("Enter an integer: ");	Asks user to enter an integer	We prompt the user to enter an integer.	5
int num = scan.nextInt();	sets the user input as "num"	We read the input integer by calling the nextInt() method because this input is an integer.	2
int num = scan.nextInt();	sets the user input as "num"	We need to read the integer that the user enters and store it in a variable.	4
scan.close();	terminates the input	We close the scanner as we do not want to process any input from the user in the rest of the program.	2
System.out.println("The integer is positivie.");	outputs "The integer is positive" if num is greater than o	This statement prints that the integer is positive.	5
System.out.println("The integer is positivie.");	outputs "The integer is positive" if num is greater than o	The printed text is followed by the end-of-line character at the end.	1
} else if ( num < 0 ) {	if num is not greater than 0, condition num less than zero is next	If the first test fails (i.e., when the integer is not positive), we need to test if the integer is negative.	4
if ( num > 0 ) {	begins an if-else condition, with num being larger than 0 as the first option	If the integer is neither positive nor negative, then we could conclude that the integer is zero.	2
if ( num > 0 ) {	begins an if-else condition, with num being larger than 0 as the first option	The conditions that tests for the integer's sign are mutually exclusive (i.e., one and only one of the conditions can be true); therefore, their order does not matter.	2
if ( num > 0 ) {	begins an if-else condition, with num being larger than 0 as the first option	To determine the sign of the integer, we need to perform two tests: one for determining whether the integer is positive and one for determining whether the integer is negative.	3
if ( num > 0 ) {	begins an if-else condition, with num being larger than 0 as the first option	Also, it is better to use if-else if statements instead of sequential if statements because an integer has only one sign and once we find the sign, we don't need to perform more tests.	2
if ( num > 0 ) {	begins an if-else condition, with num being larger than 0 as the first option	If both of these tests fail, then we could conclude that the integer is zero.	1
System.out.println("The integer is negative.");	if num is less than zero, output "The integer is negative"	The printed text is followed by the end-of-line character at the end.	1
System.out.println("The integer is negative.");	if num is less than zero, output "The integer is negative"	This statement prints that the integer is negative.	4
} else {	if the num hasn't met the first two conditions, proceed	We need to end the above if-else if statements with an else statement that its body is executed when none of the above tests are true, that is when the integer is zero.	4
System.out.println("The integer is zero.");	output "The integer is zero,"	The printed text is followed by the end-of-line character at the end.	1
System.out.println("The integer is zero.");	output "The integer is zero,"	This statement prints that the integer is zero.	5
int [] arr = { 1, 2, 3};	creates an int array with three integers defined	We initialize the array of type int to hold the specified numbers.	4
int [] arr = { 1, 2, 3};	creates an int array with three integers defined	We initialize the array by separating elements with a comma and enclosing the collection in braces { }.	3
for ( int i = 0; i < arr.length; i++ ) {	runs i from i equal to 0 the next line, adding 1 to i each time until i reaches the array length	We want to iterate over the array and increment each element in the array by 1.	3
for ( int i = 0; i < arr.length; i++ ) {	runs i from i equal to 0 the next line, adding 1 to i each time until i reaches the array length	To really change the array as we march across it, we need to use indexes so we can assign an updated value to each position as we go.	2
for ( int i = 0; i < arr.length; i++ ) {	runs i from i equal to 0 the next line, adding 1 to i each time until i reaches the array length	We need the array indexes to start at 0 (array indexes start from 0) with every integer number up to but not including the array length.	5
arr[i] += 1;	adds 1 to array position each iteration	This statement increments the element at the index i of the array by 1.	4
int num = 1234;	the number that will be printed	We need variable num to store the integer that we want to print its digits.	3
do {	loop start	We need to process the digits of the integer from right to left and print them.	1
do {	loop start	Therefore, we need to use a loop structure.	3
do {	loop start	In this program, we do this by using a do loop.	2
do {	loop start	The do loop is more appropriate than a while loop because a positive integer always has at least one digit which results in the body of the loop performing at least once.	2
System.out.println(num % 10);	number output	Each printed digit is followed by the line separator (e.g. '\n') at the end.	2
System.out.println(num % 10);	number output	We need to extract the last digit in the 1's position of the integer.	1
System.out.println(num % 10);	number output	For example, if the integer is 1234, we want to extract the digit 4 that is in 1's position of the integer.	1
System.out.println(num % 10);	number output	We do this by calculating the remainder of the division of the integer by 10.	1
System.out.println(num % 10);	number output	Then, this statement prints the last digit of the integer to the standard output stream.	3
num = num / 10;	go to the next number	Therefore, this division will remove the digit that we processed (lastDigit) and we can move on to the next digit.	3
num = num / 10;	go to the next number	We truncate the extracted digit that we processed from the original integer by dividing the integer by 10.	1
num = num / 10;	go to the next number	Note that this statement performs an integer division because both operand of the / operator are integer.	1
} while (num > 0);	loop condition	We need to check for termination conditions to avoid infinite loops.	3
} while (num > 0);	loop condition	The loop should terminate when we run out of digits to process.	2
} while (num > 0);	loop condition	We could check whether the are more digits left by checking whether the variable num, which gets updated in the body of the do loop, is greater than zero.	2
} while (num > 0);	loop condition	If variable num is greater than zero, then it must have at least one digit, and in that case, the body of the do loop will be repeated again.	2
} while (num > 0);	loop condition	The body of the while loop should repeat as long as there are more digits left that we have not processed yet.	1
int num = 1234;	sets "num" to the integer 1234	We need variable num to store the integer that we want to print its digits.	2
do {	initiates a do/while task	We need to process the digits of the integer from right to left and print them.	1
do {	initiates a do/while task	Therefore, we need to use a loop structure.	2
do {	initiates a do/while task	In this program, we do this by using a do loop.	3
do {	initiates a do/while task	The do loop is more appropriate than a while loop because a positive integer always has at least one digit which results in the body of the loop performing at least once.	2
System.out.println(num % 10);	print the remainder of num divided by 10	Each printed digit is followed by the line separator (e.g. '\n') at the end.	1
System.out.println(num % 10);	print the remainder of num divided by 10	We need to extract the last digit in the 1's position of the integer.	3
System.out.println(num % 10);	print the remainder of num divided by 10	For example, if the integer is 1234, we want to extract the digit 4 that is in 1's position of the integer.	1
System.out.println(num % 10);	print the remainder of num divided by 10	We do this by calculating the remainder of the division of the integer by 10.	4
System.out.println(num % 10);	print the remainder of num divided by 10	Then, this statement prints the last digit of the integer to the standard output stream.	3
System.out.println(num % 10);	print the remainder of num divided by 10	Each printed digit is followed by the line separator (e.g. '\n') at the end.	1
System.out.println(num % 10);	print the remainder of num divided by 10	We need to extract the last digit in the 1's position of the integer.	3
System.out.println(num % 10);	print the remainder of num divided by 10	For example, if the integer is 1234, we want to extract the digit 4 that is in 1's position of the integer.	2
System.out.println(num % 10);	print the remainder of num divided by 10	We do this by calculating the remainder of the division of the integer by 10.	4
System.out.println(num % 10);	print the remainder of num divided by 10	Then, this statement prints the last digit of the integer to the standard output stream.	3
num = num / 10;	reset num to num divided by 10	Therefore, this division will remove the digit that we processed (lastDigit) and we can move on to the next digit.	3
num = num / 10;	reset num to num divided by 10	We truncate the extracted digit that we processed from the original integer by dividing the integer by 10.	3
num = num / 10;	reset num to num divided by 10	Note that this statement performs an integer division because both operand of the / operator are integer.	2
} while (num > 0);	continue do-task while num is greater than zero	We need to check for termination conditions to avoid infinite loops.	2
} while (num > 0);	continue do-task while num is greater than zero	The loop should terminate when we run out of digits to process.	3
} while (num > 0);	continue do-task while num is greater than zero	We could check whether the are more digits left by checking whether the variable num, which gets updated in the body of the do loop, is greater than zero.	4
} while (num > 0);	continue do-task while num is greater than zero	If variable num is greater than zero, then it must have at least one digit, and in that case, the body of the do loop will be repeated again.	4
} while (num > 0);	continue do-task while num is greater than zero	The body of the while loop should repeat as long as there are more digits left that we have not processed yet.	3
Scanner scan = new Scanner(System.in);	define a new Scanner named "scan" using sytem input	To read the input values from the user, we need to define a Scanner object.	3
Scanner scan = new Scanner(System.in);	define a new Scanner named "scan" using sytem input	We need to read and process the values that the user enters.	2

Scanner scan = new Scanner(System.in);

creates a new Scanner instance which points to the input stream passed as argument.

To read the input value from the user, we need to define a Scanner object.

4

Scanner scan = new Scanner(System.in);

creates a new Scanner instance which points to the input stream passed as argument.

We need to read and process the value that the user enters.

2

Scanner scan = new Scanner(System.in);

In this case the steam is Standard input stream

To read the input value from the user, we need to define a Scanner object.

2

Scanner scan = new Scanner(System.in);

In this case the steam is Standard input stream

We need to read and process the value that the user enters.

2

System.out.println("Enter an integer for seconds: ");

prints the argument ("Enter an integer for second: ") passed to the standard console and a newline

We prompt the user to enter the seconds.

4

int seconds = scan.nextInt();

scaner object reads in a string of digits (characters) and converts them into an int type, and put them in variable entitled "seconds"

We need to read the seconds that the user enters and store it in a variable.

4

int seconds = scan.nextInt();

scaner object reads in a string of digits (characters) and converts them into an int type, and put them in variable entitled "seconds"

We read the seconds by calling the nextInt() method because this input is an integer.

4

scan.close();

closes scan the input that has been opened

We close the scanner as we do not want to process any input from the user in the rest of the program.

3

int minutes = seconds / 60;

creates varible entitled "minutes" which contain calculation of variable "second" divided by 60

To obtain the minutes in seconds, we divide the seconds by 60 because there are 60 seconds in a minute.

4

int minutes = seconds / 60;

creates varible entitled "minutes" which contain calculation of variable "second" divided by 60

Note that since both operands of division operator are integer, the fractional part of the result is truncated, if there is any.

2

int remainingSeconds = seconds % 60;

creates varible entitled "remainingSeconds" which contain calculation of variable "second" modulus by 60

This is because there are 60 seconds in a minute.

2

int remainingSeconds = seconds % 60;

creates varible entitled "remainingSeconds" which contain calculation of variable "second" modulus by 60

Note that the % operator returns the remainder of the division.

2

int remainingSeconds = seconds % 60;

creates varible entitled "remainingSeconds" which contain calculation of variable "second" modulus by 60

To obtain the remaining seconds after taking away the minutes, we have to take the remainder of the seconds divided by 60.

3

System.out.println(seconds + " seconds is " + minutes + " minutes and " + remainingSeconds + " seconds.");

prints in the console line the character passed inside the argument

This statement prints to the default standard output stream the minutes and remaining seconds from the input amount of time in seconds.

2

System.out.println(seconds + " seconds is " + minutes + " minutes and " + remainingSeconds + " seconds.");

prints in the console line the character passed inside the argument

The printed text is followed by the end-of-line character at the end.

1

int num = 15;

create variable integer entitled "num" with value 5

We define variable num to store the number that we want to find its smallest divisor.

3

int num = 15;

create variable integer entitled "num" with value 5

We could initialize it to any positive integer greater than 1.

2

int num = 15;

create variable integer entitled "num" with value 5

In this program, we initialize variable num to 15.

2

int divisor = 2;

creates variable integer entitled "division" with initial value 2

We initialize variable divisor by 2 because we want to find the smallest divisor except 1.

3

int divisor = 2;

creates variable integer entitled "division" with initial value 2

We define variable divisor to store the smallest divisor of the number.

2

int num = 15;

creates variable integer entitled "num" with initial value 5

We define variable num to store the number that we want to find its smallest divisor.

2

int num = 15;

creates variable integer entitled "num" with initial value 5

We could initialize it to any positive integer greater than 1.

2

int num = 15;

creates variable integer entitled "num" with initial value 5

In this program, we initialize variable num to 15.

2

int divisor = 2;

creates variable integer entitled "divisor" with initial value 2

We initialize variable divisor by 2 because we want to find the smallest divisor except 1.

3

int divisor = 2;

creates variable integer entitled "divisor" with initial value 2

We define variable divisor to store the smallest divisor of the number.

3

while (num % divisor != 0) {

creates looping "while" with condition while variable num modulus by the value of divisor is not null then execute the below code

Since we don't know ahead of time how many times the loop will be repeated, we need to use a while loop.

3

while (num % divisor != 0) {

creates looping "while" with condition while variable num modulus by the value of divisor is not null then execute the below code

We need to increment the divisor repeatedly as long as the divisor is not a factor of the number.

1

while (num % divisor != 0) {

creates looping "while" with condition while variable num modulus by the value of divisor is not null then execute the below code

Therefore, we need to use a loop structure.

4

while (num % divisor != 0) {

creates looping "while" with condition while variable num modulus by the value of divisor is not null then execute the below code

The condition in the while loop tests whether the body of the loop should be repeated, so it should test whether the divisor is not a factor of the number.

2

while (num % divisor != 0) {

creates looping "while" with condition while variable num modulus by the value of divisor is not null then execute the below code

We could check whether the divisor is not a factor of the number by computing the remainder of the division of the number by the divisor.

2

divisor += 1;

add divisor variable by + 1

When the divisor is not a factor of the number, we increment the variable divisor by 1.

3

System.out.println("The smallest divisor of " + num + " is " + divisor);

prints the character to the console with the argument inside it.

This statement prints to the default standard output stream the smallest divisor of the number.

2

Scanner scan = new Scanner(System.in);

defines a new scanner, named "scan" and inputs System.in

To read the input value from the user, we need to define a Scanner object.

3

Scanner scan = new Scanner(System.in);

defines a new scanner, named "scan" and inputs System.in

We need to read and process the integer that the user enters.

2

System.out.println("Enter an integer: ");

Asks user to enter an integer

We prompt the user to enter an integer.

5

int num = scan.nextInt();

sets the user input as "num"

We read the input integer by calling the nextInt() method because this input is an integer.

2

int num = scan.nextInt();

sets the user input as "num"

We need to read the integer that the user enters and store it in a variable.

4

scan.close();

terminates the input

We close the scanner as we do not want to process any input from the user in the rest of the program.

2

System.out.println("The integer is positivie.");

outputs "The integer is positive" if num is greater than o

This statement prints that the integer is positive.

5

System.out.println("The integer is positivie.");

outputs "The integer is positive" if num is greater than o

The printed text is followed by the end-of-line character at the end.

1

} else if ( num < 0 ) {

if num is not greater than 0, condition num less than zero is next

If the first test fails (i.e., when the integer is not positive), we need to test if the integer is negative.

4

if ( num > 0 ) {

begins an if-else condition, with num being larger than 0 as the first option

If the integer is neither positive nor negative, then we could conclude that the integer is zero.

2

if ( num > 0 ) {

begins an if-else condition, with num being larger than 0 as the first option

The conditions that tests for the integer's sign are mutually exclusive (i.e., one and only one of the conditions can be true); therefore, their order does not matter.

2

if ( num > 0 ) {

begins an if-else condition, with num being larger than 0 as the first option

To determine the sign of the integer, we need to perform two tests: one for determining whether the integer is positive and one for determining whether the integer is negative.

3

if ( num > 0 ) {

begins an if-else condition, with num being larger than 0 as the first option

Also, it is better to use if-else if statements instead of sequential if statements because an integer has only one sign and once we find the sign, we don't need to perform more tests.

2

if ( num > 0 ) {

begins an if-else condition, with num being larger than 0 as the first option

If both of these tests fail, then we could conclude that the integer is zero.

1

System.out.println("The integer is negative.");

if num is less than zero, output "The integer is negative"

The printed text is followed by the end-of-line character at the end.

1

System.out.println("The integer is negative.");

if num is less than zero, output "The integer is negative"

This statement prints that the integer is negative.

4

} else {

if the num hasn't met the first two conditions, proceed

We need to end the above if-else if statements with an else statement that its body is executed when none of the above tests are true, that is when the integer is zero.

4

System.out.println("The integer is zero.");

output "The integer is zero,"

The printed text is followed by the end-of-line character at the end.

1

System.out.println("The integer is zero.");

output "The integer is zero,"

This statement prints that the integer is zero.

5

int [] arr = { 1, 2, 3};

creates an int array with three integers defined

We initialize the array of type int to hold the specified numbers.

4

int [] arr = { 1, 2, 3};

creates an int array with three integers defined

We initialize the array by separating elements with a comma and enclosing the collection in braces { }.

3

for ( int i = 0; i < arr.length; i++ ) {

runs i from i equal to 0 the next line, adding 1 to i each time until i reaches the array length

We want to iterate over the array and increment each element in the array by 1.

3

for ( int i = 0; i < arr.length; i++ ) {

runs i from i equal to 0 the next line, adding 1 to i each time until i reaches the array length

To really change the array as we march across it, we need to use indexes so we can assign an updated value to each position as we go.

2

for ( int i = 0; i < arr.length; i++ ) {

runs i from i equal to 0 the next line, adding 1 to i each time until i reaches the array length

We need the array indexes to start at 0 (array indexes start from 0) with every integer number up to but not including the array length.

5

arr[i] += 1;

adds 1 to array position each iteration

This statement increments the element at the index i of the array by 1.

4

int num = 1234;

the number that will be printed

We need variable num to store the integer that we want to print its digits.

3

do {

loop start

We need to process the digits of the integer from right to left and print them.

1

do {

loop start

Therefore, we need to use a loop structure.

3

do {

loop start

In this program, we do this by using a do loop.

2

do {

loop start

The do loop is more appropriate than a while loop because a positive integer always has at least one digit which results in the body of the loop performing at least once.

2

System.out.println(num % 10);

number output

Each printed digit is followed by the line separator (e.g. '\n') at the end.

2

System.out.println(num % 10);

number output

We need to extract the last digit in the 1's position of the integer.

1

System.out.println(num % 10);

number output

For example, if the integer is 1234, we want to extract the digit 4 that is in 1's position of the integer.

1

System.out.println(num % 10);

number output

We do this by calculating the remainder of the division of the integer by 10.

1

System.out.println(num % 10);

number output

Then, this statement prints the last digit of the integer to the standard output stream.

3

num = num / 10;

go to the next number

Therefore, this division will remove the digit that we processed (lastDigit) and we can move on to the next digit.

3

num = num / 10;

go to the next number

We truncate the extracted digit that we processed from the original integer by dividing the integer by 10.

1

num = num / 10;

go to the next number

Note that this statement performs an integer division because both operand of the / operator are integer.

1

} while (num > 0);

loop condition

We need to check for termination conditions to avoid infinite loops.

3

} while (num > 0);

loop condition

The loop should terminate when we run out of digits to process.

2

} while (num > 0);

loop condition

We could check whether the are more digits left by checking whether the variable num, which gets updated in the body of the do loop, is greater than zero.

2

} while (num > 0);

loop condition

If variable num is greater than zero, then it must have at least one digit, and in that case, the body of the do loop will be repeated again.

2

} while (num > 0);

loop condition

The body of the while loop should repeat as long as there are more digits left that we have not processed yet.

1

int num = 1234;

sets "num" to the integer 1234

We need variable num to store the integer that we want to print its digits.

2

do {

initiates a do/while task

We need to process the digits of the integer from right to left and print them.

1

do {

initiates a do/while task

Therefore, we need to use a loop structure.

2

do {

initiates a do/while task

In this program, we do this by using a do loop.

3

do {

initiates a do/while task

The do loop is more appropriate than a while loop because a positive integer always has at least one digit which results in the body of the loop performing at least once.

2

System.out.println(num % 10);