Array Declaration and Initialization in DSA C

Mental Model

An array is a collection of boxes lined up, each holding a value. Declaring and initializing means setting up these boxes and putting values inside them.

Analogy: Imagine a row of mailboxes, each labeled with a number. Declaring an array is like building the mailboxes, and initializing is like putting letters inside each mailbox.

Index: 0   1   2   3   4
Array: [ ] [ ] [ ] [ ] [ ]

Dry Run Walkthrough

Input: Declare an array of 5 integers and initialize it with values 10, 20, 30, 40, 50

Goal: Create an array with 5 boxes and fill each with the given numbers

Step 1: Declare an integer array of size 5

Index: 0   1   2   3   4
Array: [ ] [ ] [ ] [ ] [ ]

Why: We need to reserve space for 5 integers

Step 2: Initialize index 0 with 10

Index: 0   1   2   3   4
Array: [10] [ ] [ ] [ ] [ ]

Why: Put the first value in the first box

Step 3: Initialize index 1 with 20

Index: 0   1   2   3   4
Array: [10] [20] [ ] [ ] [ ]

Why: Put the second value in the second box

Step 4: Initialize index 2 with 30

Index: 0   1   2   3   4
Array: [10] [20] [30] [ ] [ ]

Why: Put the third value in the third box

Step 5: Initialize index 3 with 40

Index: 0   1   2   3   4
Array: [10] [20] [30] [40] [ ]

Why: Put the fourth value in the fourth box

Step 6: Initialize index 4 with 50

Index: 0   1   2   3   4
Array: [10] [20] [30] [40] [50]

Why: Put the fifth value in the fifth box

Result:

Index: 0   1   2   3   4
Array: [10] [20] [30] [40] [50]

Annotated Code

DSA C

#include <stdio.h>

int main() {
    // Declare and initialize array with 5 integers
    int arr[5] = {10, 20, 30, 40, 50};

    // Print array elements
    for (int i = 0; i < 5; i++) {
        printf("%d ", arr[i]);
    }
    printf("\n");
    return 0;
}

int arr[5] = {10, 20, 30, 40, 50};

Declare array of size 5 and initialize each index with given values

for (int i = 0; i < 5; i++) {

Loop through each index to access array elements

printf("%d ", arr[i]);

Print the value at current index

Output

10 20 30 40 50 

Complexity Analysis

Time: O(n) because printing all n elements requires visiting each once

Space: O(n) because array stores n elements in contiguous memory

vs Alternative: Compared to dynamic allocation, static array declaration is simpler and faster but fixed size

Edge Cases

Declare array with zero size

Compilation error or no storage allocated

DSA C

int arr[0];

Initialize fewer elements than size

Remaining elements are set to zero automatically

DSA C

int arr[5] = {10, 20};

Initialize more elements than size

Compilation error due to excess initializers

DSA C

int arr[5] = {10, 20, 30, 40, 50, 60};

When to Use This Pattern

When you need to store multiple values of the same type in order, use array declaration and initialization to set up fixed-size storage.

Common Mistakes

Mistake: Forgetting to specify array size when not initializing all elements
Fix: Always specify size or initialize all elements to let compiler infer size

Mistake: Accessing array index out of declared size
Fix: Ensure index is within 0 to size-1 to avoid undefined behavior

Summary

It creates a fixed-size collection of elements stored in order.

Use it when you know the number of elements in advance and want fast access by index.

The key is that declaration reserves space and initialization fills it with values.

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