calloc function - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the calloc function do in C?

calloc allocates memory for an array of elements, initializes all bytes to zero, and returns a pointer to the allocated memory.

beginner

What is the difference between malloc and calloc?

malloc allocates memory but leaves it uninitialized. calloc allocates memory and initializes all bytes to zero.

beginner

How do you use calloc? Provide the function signature and explain parameters.

Signature: void *calloc(size_t num, size_t size);

num: number of elements to allocate.
size: size in bytes of each element.

beginner

What should you always do after using calloc to avoid memory leaks?

Always use free() to release the allocated memory when it is no longer needed.

beginner

What happens if calloc fails to allocate memory?

calloc returns NULL. You should always check the returned pointer before using it.

What does calloc(5, sizeof(int)) do?

A Frees memory allocated for 5 integers.

B Allocates memory for 5 integers but leaves it uninitialized.

C Allocates memory for 1 integer of size 5 bytes.

D Allocates memory for 5 integers and initializes all to zero.

calloc allocates memory for the number of elements (5) times the size of each element (sizeof(int)) and initializes all bytes to zero.

Which function initializes allocated memory to zero?

A calloc

B realloc

C malloc

D free

calloc initializes all allocated memory bytes to zero, unlike malloc.

What should you check after calling calloc?

A If the pointer is <code>NULL</code> to confirm allocation success.

B If the pointer is negative.

C If the pointer points to a string.

D If the pointer is an integer.

Always check if the pointer returned by calloc is NULL to ensure memory was allocated.

What happens if you forget to call free on memory allocated by calloc?

A Memory is automatically freed.

B Memory leak occurs, wasting memory.

C Program crashes immediately.

D Memory is doubled.

Not freeing allocated memory causes memory leaks, which waste system resources.

Which header file must be included to use calloc?

A <code>&lt;stdio.h&gt;</code>

B <code>&lt;string.h&gt;</code>

C <code>&lt;stdlib.h&gt;</code>

D <code>&lt;math.h&gt;</code>

calloc is declared in <stdlib.h>.

Explain how calloc works and why it might be preferred over malloc.

Think about what happens to the memory bytes after allocation.

Describe the steps you should take after calling calloc in a program.

Consider safety and resource management.

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