What is the output of the following code?public class array_output { public static void main(String args[]) { int array_variable[] = new int[10]; for (int i = 0; i < 10; ++i) { array_variable[i] = i; System.out.print(array_variable[i] + " "); i++; } }}Options: Pick one correct answer from below0 2 4 6 81 3 5 7 90 1 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 9 10
Question
What is the output of the following code?public class array_output { public static void main(String args[]) { int array_variable[] = new int[10]; for (int i = 0; i < 10; ++i) { array_variable[i] = i; System.out.print(array_variable[i] + " "); i++; } }}Options: Pick one correct answer from below0 2 4 6 81 3 5 7 90 1 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 9 10
Solution
The correct answer is 0 2 4 6 8.
Here's the step by step explanation:
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An integer array of size 10 is declared and initialized. This means the array can hold 10 integer values.
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A for loop is started with the initialization of variable i to 0. The condition for the loop to run is i < 10.
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Inside the loop, the array_variable at index i is assigned the value of i. So, for the first iteration, array_variable[0] = 0.
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The value of array_variable[i] is then printed, which is 0 for the first iteration.
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The variable i is then incremented by 1, making i = 1.
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However, at the end of the loop, i is again incremented by 1 due to the i++ in the for loop declaration. So, i becomes 2.
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The loop then continues with i = 2, and the same steps are repeated. This results in the even numbers from 0 to 8 being printed.
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The loop stops when i becomes 10, as the condition for the loop to run is i < 10.
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