calloc function - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to allocate memory for 10 integers using calloc.

C

int *arr = (int *)[1](10, sizeof(int));

Drag options to blanks, or click blank then click option'

Acalloc

Brealloc

Cfree

Dmalloc

Attempts:

3 left

💡 Hint

Common Mistakes

Using malloc instead of calloc when zero-initialization is needed.

Forgetting to cast the return value of calloc.

The calloc function allocates memory for an array of elements and initializes all bytes to zero.

2fill in blank

medium

Complete the code to check if calloc failed to allocate memory.

C

if (ptr == [1]) {
    printf("Allocation failed\n");
}

Drag options to blanks, or click blank then click option'

A1

BNULL

C0

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if pointer equals 0 instead of NULL.

Not checking the pointer at all.

When calloc fails, it returns NULL. Checking for NULL ensures the program handles allocation failure.

3fill in blank

hard

Fix the error in the calloc usage to allocate memory for 5 doubles.

C

double *data = (double *)calloc([1], sizeof(double));

Drag options to blanks, or click blank then click option'

A5

Bsizeof(double)

Csizeof(int)

D10

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the number of elements and size arguments.

Using sizeof(int) instead of sizeof(double).

The first argument to calloc is the number of elements, here 5. The second argument is the size of each element, which should be sizeof(double), but the blank is for the number of elements.

4fill in blank

hard

Fill both blanks to allocate memory for 8 characters and check if allocation succeeded.

C

char *buffer = (char *)[1]([2], sizeof(char));
if (buffer == NULL) {
    printf("Memory allocation failed\n");
}

Drag options to blanks, or click blank then click option'

Acalloc

Bmalloc

C8

D10

Attempts:

3 left

💡 Hint

Common Mistakes

Using malloc instead of calloc when zero-initialization is needed.

Not checking for NULL after allocation.

Use calloc to allocate and zero-initialize 8 characters. Then check if the pointer is NULL to detect failure.

5fill in blank

hard

Fill all three blanks to allocate memory for 12 floats, check allocation, and free the memory.

C

float *arr = (float *)[1]([2], sizeof(float));
if (arr == [3]) {
    printf("Allocation failed\n");
} else {
    free(arr);
}

Drag options to blanks, or click blank then click option'

Amalloc

B12

CNULL

Dcalloc

Attempts:

3 left

💡 Hint

Common Mistakes

Not freeing allocated memory.

Checking pointer against 0 instead of NULL.

Use calloc to allocate 12 floats. Check if the pointer is NULL to detect failure. If allocation succeeds, free the memory.

Practice

1. What does the calloc function do in C?

easy

A. Copies memory from one location to another

B. Allocates memory but leaves it uninitialized

C. Frees previously allocated memory

D. Allocates memory for multiple elements and initializes all bytes to zero

Solution

1. Step 1: Understand calloc purpose

   calloc allocates memory for an array of elements and initializes all bytes to zero.

2. Step 2: Compare with other functions

   malloc allocates memory but does not initialize it, free releases memory, and memcpy copies memory.

3. Final Answer:

   Allocates memory for multiple elements and initializes all bytes to zero -> Option D

4. Quick Check:

   calloc = zeroed memory allocation [OK]

Hint: Remember: calloc = allocate + clear (zero) memory [OK]

Common Mistakes:

• Confusing calloc with malloc (uninitialized memory)
• Thinking calloc frees memory
• Assuming calloc copies data

2. Which of the following is the correct syntax to allocate memory for 10 integers using calloc?

easy

A. int *arr = calloc(10);

B. int *arr = calloc(sizeof(int), 10);

C. int *arr = calloc(10, sizeof(int));

D. int *arr = calloc(10 * sizeof(int));

Solution

1. Step 1: Recall calloc parameters

   calloc takes two arguments: number of elements and size of each element.

2. Step 2: Match correct order and types

   int *arr = calloc(10, sizeof(int)); correctly uses calloc(10, sizeof(int)). Other options have wrong order or missing arguments.

3. Final Answer:

   int *arr = calloc(10, sizeof(int)); -> Option C

4. Quick Check:

   calloc(count, size) = correct syntax [OK]

Hint: First argument = count, second = size in calloc [OK]

Common Mistakes:

• Swapping the two arguments
• Passing only one argument
• Multiplying inside calloc instead of separate arguments

3. What will be the output of this code snippet?

int *ptr = calloc(5, sizeof(int));
for (int i = 0; i < 5; i++) {
    printf("%d ", ptr[i]);
}
free(ptr);

medium

A. Random garbage values

B. 0 0 0 0 0

C. Compilation error

D. Segmentation fault

Solution

1. Step 1: Understand calloc initialization

   calloc sets all allocated bytes to zero, so all integers start as 0.

2. Step 2: Analyze the loop output

   The loop prints each element, all zero, so output is "0 0 0 0 0 ".

3. Final Answer:

   0 0 0 0 0 -> Option B

4. Quick Check:

   calloc zeroes memory = zeros printed [OK]

Hint: calloc memory is always zeroed, so prints zeros [OK]

Common Mistakes:

• Expecting garbage values like malloc
• Forgetting to free memory (not causing output error here)
• Assuming uninitialized values

4. Identify the error in this code:

int *arr = calloc(5, sizeof(int));
arr[5] = 10;
free(arr);

medium

A. Accessing out-of-bounds index

B. Missing free call

C. Incorrect calloc syntax

D. No error, code is correct

Solution

1. Step 1: Check array indexing

   Array allocated has indices 0 to 4. Accessing arr[5] is out-of-bounds.

2. Step 2: Confirm other parts

   calloc syntax is correct and free is called properly.

3. Final Answer:

   Accessing out-of-bounds index -> Option A

4. Quick Check:

   Index 5 invalid for size 5 array [OK]

Hint: Array indices go from 0 to size-1 [OK]

Common Mistakes:

• Using index equal to size (off-by-one error)
• Forgetting to free memory
• Misunderstanding calloc parameters

5. You want to allocate memory for a 2D array of 3 rows and 4 columns using calloc. Which code snippet correctly does this?

hard

A. int (*arr)[4] = calloc(3, sizeof(*arr));

B. int **arr = calloc(3, sizeof(int*)); arr[0] = calloc(4, sizeof(int));

C. int *arr = calloc(3 * 4, sizeof(int));

D. int arr[3][4] = calloc(3, 4);

Solution

1. Step 1: Understand 2D array allocation

   int (*arr)[4] = calloc(3, sizeof(*arr)); allocates 3 rows of arrays each with 4 ints, using pointer to array syntax.

2. Step 2: Check other options

   int **arr = calloc(3, sizeof(int*)); arr[0] = calloc(4, sizeof(int)); only allocates first row's columns, not all rows. int *arr = calloc(3 * 4, sizeof(int)); allocates flat 1D array, not 2D. int arr[3][4] = calloc(3, 4); is invalid syntax.

3. Final Answer:

   int (*arr)[4] = calloc(3, sizeof(*arr)); -> Option A

4. Quick Check:

   Pointer to array + calloc rows = correct 2D allocation [OK]

Hint: Use pointer to array type for 2D calloc allocation [OK]

Common Mistakes:

• Allocating only one row's columns
• Using invalid syntax for arrays
• Allocating flat 1D array instead of 2D