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.