One-dimensional arrays in Java - Step-by-Step Execution

Concept Flow - One-dimensional arrays

Declare array variable

↓

Allocate memory for array

↓

Initialize elements

↓

Access elements by index

↓

Use elements in operations

↓

End or reuse array

This flow shows how a one-dimensional array is declared, created, filled with values, accessed by index, and used.

Execution Sample

Java

int[] numbers = new int[3];
numbers[0] = 10;
numbers[1] = 20;
numbers[2] = 30;
System.out.println(numbers[1]);

This code creates an array of 3 integers, assigns values, and prints the second element.

Execution Table

Step	Action	Array State	Index Accessed	Output
1	Declare array variable 'numbers'	null (no array yet)	N/A	N/A
2	Allocate array of size 3	[0, 0, 0]	N/A	N/A
3	Assign numbers[0] = 10	[10, 0, 0]	0	N/A
4	Assign numbers[1] = 20	[10, 20, 0]	1	N/A
5	Assign numbers[2] = 30	[10, 20, 30]	2	N/A
6	Print numbers[1]	[10, 20, 30]	1	20
7	End of program	[10, 20, 30]	N/A	N/A

💡 Program ends after printing the value at index 1.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	Final
numbers	null	[0, 0, 0]	[10, 0, 0]	[10, 20, 0]	[10, 20, 30]	[10, 20, 30]

Key Moments - 3 Insights

Why does the array start with zeros after allocation?

In Java, when you create an int array, all elements are automatically set to 0 by default, as shown in step 2 of the execution_table.

What happens if I try to access an index outside the array size?

Accessing an index outside 0 to length-1 causes an ArrayIndexOutOfBoundsException. This is not shown here but is important to avoid.

Why do we use numbers[1] to get the second element?

Array indexes start at 0 in Java, so numbers[1] accesses the second element, as shown in step 6 where the output is 20.

Visual Quiz - 3 Questions

Look at the execution_table, what is the array state after step 4?

A [10, 20, 0]

B [0, 20, 0]

C [10, 0, 0]

D [10, 20, 30]

💡 Hint

Check the 'Array State' column in row with Step 4.

At which step is the value 20 assigned to the array?

A Step 3

B Step 5

C Step 4

D Step 6

💡 Hint

Look at the 'Action' column and see when numbers[1] is assigned.

If we tried to print numbers[3], what would happen?

A Print 0

B ArrayIndexOutOfBoundsException

C Print 30

D Print null

💡 Hint

Array size is 3, valid indexes are 0,1,2 as shown in variable_tracker.

Concept Snapshot

One-dimensional arrays in Java:
- Declare: int[] arr;
- Create: arr = new int[size];
- Indexing starts at 0
- Default int values are 0
- Access with arr[index]
- Out-of-bounds causes error

Full Transcript

This visual trace shows how a one-dimensional array is declared, created with size 3, initialized with values 10, 20, 30 at indexes 0,1,2 respectively, and how accessing numbers[1] prints 20. The array starts with zeros after allocation. Indexing starts at zero, so numbers[1] is the second element. Accessing outside the array size causes an error.

Practice

1.

What is the starting index of a one-dimensional array in Java?

easy

A. 1

B. 0

C. -1

D. Depends on the array size

Solution

1. Step 1: Understand Java array indexing

   Java arrays always start indexing from 0, meaning the first element is at index 0.

2. Step 2: Confirm the starting index

   Since the first element is at index 0, the starting index is 0 regardless of array size.

3. Final Answer:

   0 -> Option B

4. Quick Check:

   Array index starts at 0 [OK]

Hint: Remember, Java arrays always start at zero index [OK]

Common Mistakes:

• Thinking arrays start at 1
• Assuming negative indexes
• Confusing with other languages

2.

Which of the following is the correct way to declare and create an integer array of size 5 in Java?

?

easy

A. int[] arr = int[5];

B. int arr = new int[5];

C. int arr[] = new int();

D. int[] arr = new int[5];

Solution

1. Step 1: Check array declaration syntax

   In Java, arrays are declared with type[], for example int[]. The new keyword creates the array with size.

2. Step 2: Validate each option

   int[] arr = new int[5]; uses correct syntax: int[] arr = new int[5]; Options A, B, and C have syntax errors or missing parts.

3. Final Answer:

   int[] arr = new int[5]; -> Option D

4. Quick Check:

   Correct array declaration uses new int[size] [OK]

Hint: Use 'type[] name = new type[size];' for arrays [OK]

Common Mistakes:

• Missing new keyword
• Wrong brackets placement
• Using parentheses instead of brackets

3.

What will be the output of the following Java code?

int[] numbers = {10, 20, 30, 40};
System.out.println(numbers[2]);

medium

A. 40

B. 20

C. 30

D. ArrayIndexOutOfBoundsException

Solution

1. Step 1: Identify the array elements and indexes

   The array 'numbers' has elements: index 0=10, 1=20, 2=30, 3=40.

2. Step 2: Access the element at index 2

   numbers[2] refers to the third element, which is 30.

3. Final Answer:

   30 -> Option C

4. Quick Check:

   Index 2 value is 30 [OK]

Hint: Remember array index starts at 0, so 2 means third element [OK]

Common Mistakes:

• Confusing index 2 with 1 or 3
• Expecting 40 instead of 30
• Thinking index starts at 1

4.

Find the error in the following code snippet:

int[] arr = new int[3];
arr[3] = 10;
System.out.println(arr[3]);

medium

A. Index 3 is out of bounds for array of size 3

B. Missing initialization of array elements

C. Array size should be 4

D. Syntax error in array declaration

Solution

1. Step 1: Check array size and valid indexes

   The array size is 3, so valid indexes are 0, 1, and 2.

2. Step 2: Identify the invalid index usage

   Accessing arr[3] is invalid and causes ArrayIndexOutOfBoundsException.

3. Final Answer:

   Index 3 is out of bounds for array of size 3 -> Option A

4. Quick Check:

   Array indexes go from 0 to size-1 [OK]

Hint: Array indexes must be less than size [OK]

Common Mistakes:

• Using index equal to size
• Assuming array auto-expands
• Ignoring runtime exceptions

5.

Given an integer array arr of size 5 initialized as {2, 4, 6, 8, 10}, which code snippet correctly doubles each element's value?

?

hard

A. for(int i = 0; i < arr.length; i++) { arr[i] *= 2; }

B. for(int i = 0; i <= arr.length; i++) { arr[i] = arr[i] * 2; }

C. for(int i = 1; i < arr.length; i++) { arr[i] = arr[i] * 2; }

D. for(int i = 0; i < arr.length - 1; i++) { arr[i] = arr[i] * 2; }

Solution

1. Step 1: Understand array length and loop bounds

   The array length is 5, so valid indexes are 0 to 4. Loop must run from 0 to less than length.

2. Step 2: Evaluate each option's loop condition

   for(int i = 0; i <= arr.length; i++) { arr[i] = arr[i] * 2; } uses i <= arr.length which causes out of bounds error at i=5. for(int i = 0; i < arr.length; i++) { arr[i] *= 2; } correctly uses i < arr.length. for(int i = 1; i < arr.length; i++) { arr[i] = arr[i] * 2; } starts at 1, missing index 0. for(int i = 0; i < arr.length - 1; i++) { arr[i] = arr[i] * 2; } stops at length-1, missing last element.

3. Final Answer:

   for(int i = 0; i < arr.length; i++) { arr[i] *= 2; } -> Option A

4. Quick Check:

   Loop from 0 to length-1 doubles all elements [OK]

Hint: Use i < arr.length to avoid index errors [OK]

Common Mistakes:

• Using <= arr.length causing errors
• Starting loop at 1 missing first element
• Stopping loop early missing last element