Common pointer errors - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of pointer arithmetic

What is the output of this C code snippet?

C

#include <stdio.h>

int main() {
    int arr[] = {10, 20, 30};
    int *p = arr;
    p++;
    printf("%d\n", *p);
    return 0;
}

A 30

B 20

C 10

D Compilation error

💡 Hint

Remember that incrementing a pointer moves it to the next element in the array.

Initially, p points to arr[0] which is 10. After p++, it points to arr[1] which is 20. So, *p prints 20.

❓ Predict Output

intermediate

Dereferencing a NULL pointer

What happens when this code runs?

C

#include <stdio.h>

int main() {
    int *p = NULL;
    printf("%d\n", *p);
    return 0;
}

A Runtime error (segmentation fault)

B Prints garbage value

C Prints 0

D Compilation error

💡 Hint

Think about what happens when you try to access memory through a NULL pointer.

Dereferencing a NULL pointer causes a runtime error (segmentation fault) because it points to no valid memory.

🔧 Debug

advanced

Identify the pointer error

What is the error in this code?

C

#include <stdio.h>

int main() {
    int *p;
    *p = 5;
    printf("%d\n", *p);
    return 0;
}

A Incorrect format specifier in <code>printf</code>

B Missing <code>return</code> statement

C Pointer <code>p</code> is uninitialized and used to write memory

D Array index out of bounds

💡 Hint

Think about what p points to before assignment.

The pointer p is declared but not initialized to point to valid memory. Writing to *p causes undefined behavior.

📝 Syntax

advanced

Syntax error with pointer declaration

Which option contains the correct pointer declaration?

A int *p;

B int * p[];

C int p*;

D int &p;

💡 Hint

Remember how to declare a pointer to an int in C.

int *p; declares p as a pointer to an int. Option A uses C++ reference syntax, C does not support it. Option A is invalid syntax. Option A declares an array of pointers, not a single pointer.

🚀 Application

expert

Pointer and memory allocation behavior

What is the output of this program?

C

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

int main() {
    int *p = malloc(sizeof(int));
    *p = 42;
    free(p);
    printf("%d\n", *p);
    return 0;
}

A 42

B 0

C Compilation error

D Runtime error (use after free)

💡 Hint

Consider what happens when you access memory after it has been freed.

After free(p), the memory is released. Accessing *p is undefined behavior and usually causes a runtime error (use after free).