Array Declaration and Initialization in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of partially initialized array

What is the output of the following C code snippet?

DSA C

int arr[5] = {1, 2};
for (int i = 0; i < 5; i++) {
    printf("%d ", arr[i]);
}

A 1 2 0 0 1

B 1 2 3 4 5

C 1 2 0 0 0

D 1 2 2 2 2

💡 Hint

Uninitialized elements in a partially initialized array are set to zero.

In C, when you initialize an array with fewer elements than its size, the remaining elements are set to zero.

So arr[0] = 1, arr[1] = 2, and arr[2], arr[3], arr[4] = 0.

❓ Predict Output

intermediate

Output of array initialization with explicit size and values

What is the output of this C code?

DSA C

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

A 5 10 15

B 5 10 15 0

C 0 0 0

D 15 10 5

💡 Hint

All elements are explicitly initialized.

The array has size 3 and all elements are initialized explicitly.

So the output prints 5, 10, and 15 in order.

❓ Predict Output

advanced

Output of array without explicit size but with initialization

What will be printed by this C code?

DSA C

int arr[] = {7, 8, 9, 10};
for (int i = 0; i < 4; i++) {
    printf("%d ", arr[i]);
}

A 7 8 9 10

B 7 8 9 10 0

C 0 0 0 0

D 10 9 8 7

💡 Hint

Array size is inferred from initializer list.

When size is omitted, the compiler sets the array size to the number of initializer elements.

So arr has size 4 with values 7,8,9,10.

❓ Predict Output

advanced

Output of multi-dimensional array initialization

What is the output of this C code snippet?

DSA C

int arr[2][3] = {{1, 2, 3}, {4}};
for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 3; j++) {
        printf("%d ", arr[i][j]);
    }
}

A 1 2 3 4 5 6

B 1 2 3 4 0 0

C 1 2 3 4 4 4

D 1 2 3 0 0 0

💡 Hint

Uninitialized elements in inner arrays are zero.

The first row is fully initialized: 1, 2, 3.

The second row has only the first element initialized to 4; the rest are zero.

🧠 Conceptual

expert

Size of array with partial initialization and no explicit size

Consider the declaration:
int arr[] = {3, 6, 9};
What is the size of the array arr in memory?

A Depends on compiler

B 0

C Size must be explicitly specified

D 3

💡 Hint

When size is omitted, it is inferred from initializer list length.

The compiler counts the number of elements in the initializer list and sets the array size accordingly.

Here, 3 elements means size 3.

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