calloc function - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this calloc usage?

Consider the following C code snippet using calloc. What will be printed?

C

#include <stdio.h>
#include <stdlib.h>

int main() {
    int *arr = (int *)calloc(5, sizeof(int));
    if (!arr) return 1;
    for (int i = 0; i < 5; i++) {
        printf("%d ", arr[i]);
    }
    free(arr);
    return 0;
}

A 0 0 0 0 0

B Random garbage values printed

C Compilation error due to missing header

D Segmentation fault at runtime

💡 Hint

calloc initializes allocated memory to zero.

calloc allocates memory and sets all bytes to zero, so all integers in the array are zero.

❓ Predict Output

intermediate

What happens if calloc is called with zero elements?

What will be the output or behavior of this code?

C

#include <stdio.h>
#include <stdlib.h>

int main() {
    int *ptr = (int *)calloc(0, sizeof(int));
    if (ptr == NULL) {
        printf("NULL pointer\n");
    } else {
        printf("Non-NULL pointer\n");
    }
    free(ptr);
    return 0;
}

A NULL pointer

B Runtime crash

C Compilation error

D Non-NULL pointer

💡 Hint

calloc with zero elements may return a unique pointer that should not be dereferenced.

calloc(0, size) may return a non-NULL pointer that should not be dereferenced but can be freed safely.

🔧 Debug

advanced

Why does this calloc usage cause a runtime error?

Examine the code below. It compiles but crashes at runtime. What is the cause?

C

#include <stdio.h>
#include <stdlib.h>

int main() {
    int *arr = calloc(5, sizeof(int));
    for (int i = 0; i <= 5; i++) {
        arr[i] = i * 2;
    }
    for (int i = 0; i < 5; i++) {
        printf("%d ", arr[i]);
    }
    free(arr);
    return 0;
}

A calloc was not cast to int* causing type mismatch

B The printf format specifier is incorrect

C The loop writes out of bounds at index 5 causing undefined behavior

D Memory was not freed before program ends

💡 Hint

Check the loop boundary conditions carefully.

The loop uses i <= 5 which accesses arr[5], but valid indices are 0 to 4, causing out-of-bounds write.

🧠 Conceptual

advanced

What is the main difference between malloc and calloc?

Choose the correct statement about calloc compared to malloc.

A calloc allocates memory faster than malloc always

B calloc allocates and initializes memory to zero, malloc only allocates without initialization

C malloc allocates memory and initializes it to zero, calloc does not initialize

D malloc returns NULL on failure, calloc aborts the program

💡 Hint

Think about what happens to the memory content after allocation.

calloc sets all allocated bytes to zero, while malloc leaves memory uninitialized (contains garbage).

❓ Predict Output

expert

What is the output of this calloc and pointer arithmetic code?

Analyze the code and select the correct output.

C

#include <stdio.h>
#include <stdlib.h>

int main() {
    int *p = calloc(3, sizeof(int));
    if (!p) return 1;
    *(p + 1) = 10;
    p++;
    printf("%d\n", *p);
    free(p - 1);
    return 0;
}

A 10

B 0

C Undefined behavior due to invalid free

D Compilation error due to pointer arithmetic

💡 Hint

Remember how pointer arithmetic and calloc initialization work.

calloc initializes all to zero, then *(p+1) = 10 sets second element. p++ moves pointer to second element, so *p is 10. free(p-1) frees original pointer.

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