Array initialization in C

Introduction

Arrays hold many values of the same type together. Initializing an array means giving it starting values so it is ready to use.

When you want to store a list of numbers or characters to use later.

When you know the values in advance and want to set them at the start.

When you want to create an empty array with default values before filling it.

When you want to avoid garbage or random values in your array.

When you want to quickly set all elements to zero or another value.

Syntax

C

/* Define and initialize an array with values */
int numbers[5] = {1, 2, 3, 4, 5};

/* Define an array and initialize all elements to zero */
int zeros[5] = {0};

/* Define an array without initialization (values are garbage) */
int uninitialized[5];

You must specify the array size or provide enough values to infer it.

If you provide fewer values than the size, remaining elements are set to zero.

Examples

Array of size 3 initialized with three values.

C

int scores[3] = {10, 20, 30};

Array of size 4 initialized with all zeros.

C

int empty[4] = {0};

Array of size 5, first two elements set, rest are zero.

C

int partial[5] = {1, 2};

Array declared without initialization, contains garbage values.

C

int no_init[3];

Sample Program

This program shows four arrays: fully initialized, zero initialized, partially initialized, and uninitialized. It prints their contents to see the difference.

C

#include <stdio.h>

int main() {
    int numbers[5] = {5, 10, 15, 20, 25};
    int zeros[5] = {0};
    int partial[5] = {1, 2};
    int uninitialized[5];

    printf("numbers array: ");
    for (int i = 0; i < 5; i++) {
        printf("%d ", numbers[i]);
    }
    printf("\n");

    printf("zeros array: ");
    for (int i = 0; i < 5; i++) {
        printf("%d ", zeros[i]);
    }
    printf("\n");

    printf("partial array: ");
    for (int i = 0; i < 5; i++) {
        printf("%d ", partial[i]);
    }
    printf("\n");

    printf("uninitialized array (values may vary): ");
    for (int i = 0; i < 5; i++) {
        printf("%d ", uninitialized[i]);
    }
    printf("\n");

    return 0;
}

Important Notes

Time complexity to initialize is O(n) where n is array size.

Space complexity is O(n) for the array storage.

Uninitialized arrays contain garbage values and can cause bugs if used before setting.

Use zero initialization to avoid unpredictable values.

Partial initialization sets remaining elements to zero automatically.

Summary

Arrays store multiple values of the same type in a fixed size.

Initialization sets starting values to avoid garbage data.

You can fully, partially, or zero initialize arrays in C.

Practice

1. Which of the following correctly declares and initializes an integer array of size 3 with values 1, 2, and 3 in C?

easy

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

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

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

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

Solution

1. Step 1: Understand array declaration syntax

   In C, arrays are declared with a type, name, and size in square brackets, e.g., int arr[3].

2. Step 2: Check initialization format

   Initialization uses curly braces with comma-separated values, e.g., {1, 2, 3}.

3. Final Answer:

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

4. Quick Check:

   Correct array declaration and initialization [OK]

Hint: Use curly braces {} with size in brackets [] for arrays [OK]

Common Mistakes:

• Missing square brackets for size
• Using parentheses instead of braces
• Assigning array without braces

2. Which of the following is a syntax error when initializing an array in C?

easy

A. int nums[4] = {10, 20, 30, 40};

B. int nums[5] = {0};

C. int nums[] = {5, 6, 7};

D. int nums[3] = {1, 2, 3, 4};

Solution

1. Step 1: Check array size and initializer count

   Array size is 3 but initializer has 4 values, which is invalid in C.

2. Step 2: Verify other options

   int nums[4] = {10, 20, 30, 40}; matches size and values, C infers size from initializer, B zero-initializes first element and rest are zero.

3. Final Answer:

   int nums[3] = {1, 2, 3, 4}; -> Option D

4. Quick Check:

   Initializer count > array size causes syntax error [OK]

Hint: Initializer list cannot exceed declared array size [OK]

Common Mistakes:

• Providing more initial values than array size
• Omitting size but giving initializer with extra values
• Using wrong braces

3. What is the output of the following C code?

int arr[5] = {1, 2};
printf("%d %d %d %d %d", arr[0], arr[1], arr[2], arr[3], arr[4]);

medium

A. 1 2 0 0 0

B. 1 2 garbage garbage garbage

C. 1 2 3 4 5

D. Compilation error

Solution

1. Step 1: Understand partial initialization

   When fewer values are given, remaining elements are set to zero automatically.

2. Step 2: Check printed values

   arr[0] = 1, arr[1] = 2, arr[2], arr[3], arr[4] = 0 each.

3. Final Answer:

   1 2 0 0 0 -> Option A

4. Quick Check:

   Partial init fills rest with zero [OK]

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

Common Mistakes:

• Assuming uninitialized elements have garbage values
• Expecting compilation error for partial init
• Confusing array size with initializer count

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

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

medium

A. Missing semicolon after array declaration.

B. Array size must be omitted when initializing.

C. Arrays cannot be assigned values after declaration like this.

D. Braces should be parentheses for assignment.

Solution

1. Step 1: Understand array assignment rules

   In C, arrays cannot be assigned values after declaration using = and braces.

2. Step 2: Check other options

   Semicolon is present, size can be specified, braces are correct for initialization but not assignment.

3. Final Answer:

   Arrays cannot be assigned values after declaration like this. -> Option C

4. Quick Check:

   Array assignment after declaration is invalid [OK]

Hint: Arrays must be initialized at declaration, not assigned later [OK]

Common Mistakes:

• Trying to assign array with braces after declaration
• Confusing array assignment with pointer assignment
• Using parentheses instead of braces

5. You want to create an integer array of size 6 where the first two elements are 10 and 20, and the rest are zeros. Which of the following is the best way to initialize it in C?

hard

A. int arr[6] = {0}; arr[0] = 10; arr[1] = 20;

B. int arr[6] = {10, 20, 0, 0, 0, 0};

C. int arr[6] = {10, 20};

D. int arr[6]; arr[0] = 10; arr[1] = 20;

Solution

1. Step 1: Analyze partial initialization

   int arr[6] = {10, 20}; partially initializes first two elements; rest are zero by default, so it works.

2. Step 2: Compare with explicit zero initialization and assignment

   int arr[6] = {0}; arr[0] = 10; arr[1] = 20; explicitly zeroes all elements then assigns first two, ensuring clarity and no garbage.

3. Step 3: Evaluate best practice

   int arr[6] = {0}; arr[0] = 10; arr[1] = 20; is best for clarity and safety; int arr[6] = {10, 20}; works but less explicit; int arr[6] = {10, 20, 0, 0, 0, 0}; is verbose; int arr[6]; arr[0] = 10; arr[1] = 20; leaves uninitialized elements.

4. Final Answer:

   int arr[6] = {0}; arr[0] = 10; arr[1] = 20; -> Option A

5. Quick Check:

   Explicit zero init then assign first elements [OK]

Hint: Zero initialize first, then assign specific elements for clarity [OK]

Common Mistakes:

• Assuming partial init always safe without explicit zero
• Not initializing array before assignment
• Overwriting zeros with uninitialized values