Array Declaration and Initialization in DSA C - Execution Trace

Concept Flow - Array Declaration and Initialization

Declare array variable

↓

Allocate memory for array elements

↓

Initialize elements with values

↓

Array ready for use

The flow shows declaring an array variable, allocating space for elements, initializing them, and then the array is ready.

Execution Sample

DSA C

int arr[5] = {1, 2, 3, 4, 5};

Declares an integer array of size 5 and initializes it with values 1 to 5.

Execution Table

Step	Operation	Array Size	Elements Initialized	Visual State
1	Declare array variable 'arr'	5	None	[_][_][_][_][_]
2	Allocate memory for 5 elements	5	None	[_][_][_][_][_]
3	Initialize arr[0] = 1	5	1	[1][_][_][_][_]
4	Initialize arr[1] = 2	5	2	[1][2][_][_][_]
5	Initialize arr[2] = 3	5	3	[1][2][3][_][_]
6	Initialize arr[3] = 4	5	4	[1][2][3][4][_]
7	Initialize arr[4] = 5	5	5	[1][2][3][4][5]
8	Array fully initialized	5	5	[1][2][3][4][5]

💡 All 5 elements initialized, array ready for use.

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	Final
arr[0]	undefined	1	1	1	1	1	1
arr[1]	undefined	undefined	2	2	2	2	2
arr[2]	undefined	undefined	undefined	3	3	3	3
arr[3]	undefined	undefined	undefined	undefined	4	4	4
arr[4]	undefined	undefined	undefined	undefined	undefined	5	5

Key Moments - 3 Insights

Why do array elements start as undefined before initialization?

Before initialization (see steps 1 and 2 in execution_table), memory is allocated but not set, so elements hold garbage or undefined values.

What happens if we initialize fewer elements than the array size?

Only the specified elements get values; the rest remain undefined or zero depending on language rules. Here, all 5 are initialized (steps 3-7).

Why is the array size fixed at declaration?

In C, array size must be known at compile time to allocate fixed memory space, shown in step 1 and 2.

Visual Quiz - 3 Questions

Look at the execution_table, what is the visual state after step 5?

A [1][2][_][_][_]

B [1][2][3][4][_]

C [1][2][3][_][_]

D [1][2][3][4][5]

💡 Hint

Check the 'Visual State' column for step 5 in execution_table.

At which step does the array become fully initialized?

A Step 5

B Step 8

C Step 7

D Step 6

💡 Hint

Look for the row with 'Array fully initialized' in the Operation column.

If we declared int arr[3] = {1, 2}; how would the visual state after initialization change?

A [1][2][0]

B [1][2][3]

C [1][2][_]

D [1][2][undefined]

💡 Hint

In C, missing elements in initialization default to zero.

Concept Snapshot

Array Declaration and Initialization in C:
- Declare with fixed size: int arr[5];
- Allocate memory for all elements at once
- Initialize elements with values: int arr[5] = {1,2,3,4,5};
- Uninitialized elements hold garbage or zero if partially initialized
- Array size fixed at compile time

Full Transcript

This concept shows how to declare and initialize an array in C. First, you declare the array with a fixed size. Then memory is allocated for all elements. Next, each element is initialized with a value. The array is then ready to use with all elements set. If fewer elements are initialized, the rest default to zero. The array size cannot change after declaration.

Practice

1. Which of the following is the correct way to declare an integer array of size 5 in C?

easy

A. array int arr[5];

B. int arr;

C. int arr = 5;

D. int arr[5];

Solution

1. Step 1: Understand array declaration syntax in C

   In C, to declare an array of integers with size 5, the syntax is int arr[5];.

2. Step 2: Check each option

   int arr[5]; matches the correct syntax. Options A, B, and D are incorrect because they either miss the size, use wrong keywords, or assign a value instead of declaring an array.

3. Final Answer:

   int arr[5]; -> Option D

4. Quick Check:

   Array declaration = int arr[5]; [OK]

Hint: Array size goes inside square brackets after type [OK]

Common Mistakes:

• Forgetting to specify size inside brackets
• Using wrong keywords like 'array'
• Assigning a value instead of declaring

2. Which of the following is the correct way to initialize an integer array with values 1, 2, 3 in C?

easy

A. int arr[3] = (1, 2, 3);

B. int arr = {1, 2, 3};

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

D. int arr[3] = [1, 2, 3];

Solution

1. Step 1: Recall array initialization syntax

   In C, arrays are initialized using curly braces with values inside, and size must be specified if not inferred.

2. Step 2: Evaluate each option

   int arr[3] = {1, 2, 3}; correctly declares an array of size 3 and initializes it with values 1, 2, 3. int arr = {1, 2, 3}; misses the size, C uses parentheses instead of braces, and D uses square brackets which is invalid.

3. Final Answer:

   int arr[3] = {1, 2, 3}; -> Option C

4. Quick Check:

   Array init uses braces and size = int arr[3] = {1, 2, 3}; [OK]

Hint: Use curly braces {} to initialize arrays with size [OK]

Common Mistakes:

• Using parentheses () instead of braces {}
• Omitting array size when required
• Using square brackets [] for initialization

3. What will be the output of the following C code?

int arr[4] = {10, 20};
for(int i = 0; i < 4; i++) {
    printf("%d ", arr[i]);
}

medium

A. 10 20 0 0

B. 10 20 0 20

C. 10 20 10 20

D. 10 20 garbage garbage

Solution

1. Step 1: Understand partial initialization of arrays

   When an array is partially initialized in C, the remaining elements are set to zero automatically.

2. Step 2: Trace the loop output

   arr[0] = 10, arr[1] = 20, arr[2] = 0, arr[3] = 0. The loop prints all four elements separated by spaces.

3. Final Answer:

   10 20 0 0 -> Option A

4. Quick Check:

   Partial init fills rest with 0 = 10 20 0 0 [OK]

Hint: Uninitialized array elements default to 0 if partially initialized [OK]

Common Mistakes:

• Assuming uninitialized elements have garbage values
• Confusing partial initialization with full initialization
• Miscounting array size in loop

4. Identify the error in the following C code snippet:

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

medium

A. Array size is not specified

B. Cannot assign values to array after declaration using braces

C. Missing semicolon after array declaration

D. Array elements must be initialized with parentheses

Solution

1. Step 1: Understand array assignment rules in C

   In C, arrays cannot be assigned values using braces after declaration. Initialization must happen at declaration.

2. Step 2: Check the code snippet

   The code tries to assign values to arr using braces after declaration, which is invalid syntax and causes a compilation error.

3. Final Answer:

   Cannot assign values to array after declaration using braces -> Option B

4. Quick Check:

   Array assignment after declaration with {} is invalid [OK]

Hint: Initialize arrays only at declaration with braces [OK]

Common Mistakes:

• Trying to assign arrays after declaration with {}
• Confusing initialization with assignment
• Using parentheses instead of braces

5. Given the following code, what will be the output?

int arr[] = {5, 10, 15};
int *ptr = arr;
for(int i = 0; i < 3; i++) {
    printf("%d ", *(ptr + i));
}

Choose the correct output.

hard

A. 5 10 15

B. 5 0 15

C. Garbage values

D. Compilation error

Solution

1. Step 1: Understand pointer and array relationship

   In C, the name of the array is a pointer to its first element. Pointer arithmetic allows accessing elements by offset.

2. Step 2: Trace the loop output

   Pointer ptr points to arr[0]. *(ptr + i) accesses arr[i]. Loop prints 5, 10, 15 separated by spaces.

3. Final Answer:

   5 10 15 -> Option A

4. Quick Check:

   Pointer arithmetic accesses array elements correctly = 5 10 15 [OK]

Hint: Array name acts as pointer to first element [OK]

Common Mistakes:

• Confusing pointer arithmetic syntax
• Assuming pointer points to garbage
• Forgetting array size in loop