Why pointers are needed in C - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a pointer to an integer.

C

int [1] ptr;

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

A#

B&

C*

D%

Attempts:

3 left

💡 Hint

Common Mistakes

Using & instead of * to declare a pointer.

Forgetting the * symbol when declaring a pointer.

The asterisk (*) is used to declare a pointer variable in C.

2fill in blank

medium

Complete the code to assign the address of variable 'a' to pointer 'ptr'.

C

int a = 10;
int *ptr;
ptr = [1]a;

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

A%

B*

C#

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Using * instead of & to get the address.

Assigning the variable value instead of its address.

The ampersand (&) operator gives the address of a variable in C.

3fill in blank

hard

Fix the error in the code to correctly print the value pointed by 'ptr'.

C

int a = 5;
int *ptr = &a;
printf("%d", [1]);

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

A*ptr

Bptr

C&ptr

D&&ptr

Attempts:

3 left

💡 Hint

Common Mistakes

Printing the pointer address instead of the value.

Using &ptr which gives the address of the pointer itself.

Using *ptr accesses the value stored at the address the pointer points to.

4fill in blank

hard

Fill both blanks to create a pointer to an integer and assign it the address of variable 'num'.

C

int num = 20;
[1] ptr;
ptr = [2]num;

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

Aint *

Bint

C&

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Declaring pointer without *.

Assigning value instead of address.

To declare a pointer to int, use 'int *'. To assign address, use '&'.

5fill in blank

hard

Fill all three blanks to create a pointer, assign it the address of 'val', and print the value using the pointer.

C

int val = 30;
[1] ptr;
ptr = [2]val;
printf("%d", [3]ptr);

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

Aint *

B&

C*

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting * in declaration.

Assigning value instead of address.

Printing pointer address instead of value.

Declare pointer with 'int *', assign address with '&', and dereference with '*'.

Practice

1. Why do we need pointers in C programming?

easy

A. To store large amounts of data directly

B. To avoid using variables altogether

C. To replace all variables with memory addresses

D. To allow functions to modify variables outside their own scope

Solution

1. Step 1: Understand the role of pointers in functions

   Pointers hold addresses, so functions can access and change variables outside their local scope.

2. Step 2: Compare other options

   Options A, C, and D misunderstand pointers' purpose; pointers do not store large data directly or replace variables entirely.

3. Final Answer:

   To allow functions to modify variables outside their own scope -> Option D

4. Quick Check:

   Pointers enable external variable modification = B [OK]

Hint: Pointers let functions change outside variables [OK]

Common Mistakes:

• Thinking pointers store actual data, not addresses
• Believing pointers replace variables completely
• Confusing pointers with regular variables

2. Which of the following is the correct way to declare a pointer to an integer in C?

easy

A. *int ptr;

B. int *ptr;

C. int ptr*;

D. pointer int ptr;

Solution

1. Step 1: Recall pointer declaration syntax

   In C, a pointer to int is declared as int *ptr; where * indicates pointer type.

2. Step 2: Check other options for syntax errors

   Options A, B, and D do not follow correct C syntax for pointer declaration.

3. Final Answer:

   int *ptr; -> Option B

4. Quick Check:

   Pointer to int declared as int *ptr [OK]

Hint: Pointer declaration: type *name; [OK]

Common Mistakes:

• Placing * after variable name instead of type
• Using invalid keywords like 'pointer'
• Misplacing the asterisk symbol

3. What will be the output of the following C code?

void update(int *p) {
    *p = 10;
}

int main() {
    int x = 5;
    update(&x);
    printf("%d", x);
    return 0;
}

medium

A. 10

B. 5

C. Address of x

D. Compilation error

Solution

1. Step 1: Understand pointer usage in function

   The function update takes a pointer to int and sets the value at that address to 10.

2. Step 2: Trace the main function execution

   Main declares x=5, passes its address to update, which changes x to 10. Then printf prints 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Pointer changes variable value = 10 [OK]

Hint: Pointer lets function change original variable [OK]

Common Mistakes:

• Thinking the value remains 5 after function call
• Confusing pointer address with value
• Expecting compilation error due to pointer usage

4. Identify the error in the following code snippet:

void change(int *p) {
    p = 20;
}

int main() {
    int x = 5;
    change(&x);
    printf("%d", x);
    return 0;
}

medium

A. Pointer p is assigned a value instead of the pointed variable

B. Missing & operator when calling change function

C. printf format specifier is incorrect

D. Variable x is not initialized

Solution

1. Step 1: Analyze pointer assignment inside function

   The code assigns 20 to pointer variable p itself, not to the value it points to. It should be *p = 20;.

2. Step 2: Check function call and printf

   The function call passes address correctly, and printf uses correct format specifier, so no errors there.

3. Final Answer:

   Pointer p is assigned a value instead of the pointed variable -> Option A

4. Quick Check:

   Assign *p, not p, to change value [OK]

Hint: Assign to *p, not p, to change pointed value [OK]

Common Mistakes:

• Assigning value to pointer variable instead of dereferenced pointer
• Forgetting to use * to access pointed data
• Assuming pointer changes original variable without dereference

5. You want to write a function in C that swaps the values of two integers. Why do you need pointers for this task?

hard

A. Because pointers prevent variables from being changed

B. Because pointers automatically swap values without extra code

C. Because passing variables directly copies values, so changes won't affect originals

D. Because pointers store the sum of two integers

Solution

1. Step 1: Understand parameter passing in C

   When variables are passed directly, only copies are sent, so changes inside function don't affect originals.

2. Step 2: Role of pointers in swapping

   Pointers allow passing addresses, so the function can access and swap the original variables' values.

3. Final Answer:

   Because passing variables directly copies values, so changes won't affect originals -> Option C

4. Quick Check:

   Pointers let function modify original variables [OK]

Hint: Swap needs pointers to change original variables [OK]

Common Mistakes:

• Thinking pointers swap automatically without code
• Confusing pointers with arithmetic operations
• Believing pointers prevent changes instead of enabling them