Pointer declaration - Step-by-Step Execution

Concept Flow - Pointer declaration

Start

↓

Declare variable

↓

Declare pointer with *

↓

Pointer stores address of variable

↓

Use pointer to access variable

↓

End

This flow shows how a pointer is declared to store the address of a variable and then used to access that variable.

Execution Sample

C

int x = 10;
int *p = &x;
printf("Value via pointer: %d", *p);

Declare an integer x, declare a pointer p to int, assign p the address of x, then print the value pointed by p.

Execution Table

Step	Action	Variable	Pointer	Value Accessed	Output
1	Declare int x = 10	x=10	p=uninitialized	-	-
2	Declare pointer p to int	x=10	p=uninitialized	-	-
3	Assign p = &x (address of x)	x=10	p=&x	-	-
4	Access value via *p	x=10	p=&x	10	Value via pointer: 10

💡 Program ends after printing value accessed through pointer *p.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4
x	undefined	10	10	10	10
p	undefined	undefined	undefined	&x	&x

Key Moments - 3 Insights

Why do we use * in the pointer declaration?

The * tells the compiler that p is a pointer to an int, not an int itself. See execution_table step 2 where p is declared as a pointer.

What does &x mean in the assignment to p?

&x means the address of variable x. In step 3, p stores this address, so p points to x.

How does *p give the value of x?

*p means 'the value at the address stored in p'. Since p points to x, *p accesses x's value, shown in step 4.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what does p hold?

A The value 10

B The address of x

C Undefined

D The value of *p

💡 Hint

Check the 'Pointer' column at step 3 in the execution_table.

At which step is the value 10 accessed through the pointer?

A Step 4

B Step 1

C Step 3

D Step 2

💡 Hint

Look at the 'Value Accessed' and 'Output' columns in the execution_table.

If we change int *p = &x; to int *p = NULL;, what would *p access?

A Value 10

B Address of x

C Undefined or error

D Value 0

💡 Hint

Refer to pointer initialization and what happens if pointer does not point to a valid address.

Concept Snapshot

Pointer declaration in C:
int *p; declares p as a pointer to int.
Use & to get variable address: p = &x;
Use *p to access value at that address.
Pointers store addresses, not values.
Dereferencing *p accesses the variable pointed to.

Full Transcript

This example shows how to declare a pointer in C. First, an integer variable x is declared and set to 10. Then, a pointer p is declared with int *p, meaning p will hold the address of an int. Next, p is assigned the address of x using &x. Finally, the value of x is accessed by dereferencing p with *p, which prints 10. The execution table traces each step, showing variable and pointer states. Key points include understanding that * in declaration means pointer type, & gets address, and *p accesses the value at that address.

Practice

1. What does the following declaration mean in C?
int *ptr;

easy

A. ptr is a pointer to a character variable

B. ptr is a pointer to an integer variable

C. ptr is an integer variable

D. ptr is a function returning an integer

Solution

1. Step 1: Understand the * symbol in declaration

   The * before ptr means ptr is a pointer variable, not a normal integer.

2. Step 2: Check the type before *

   The type before * is int, so ptr points to an integer variable.

3. Final Answer:

   ptr is a pointer to an integer variable -> Option B

4. Quick Check:

   int *ptr means pointer to int [OK]

Hint: Look for * before variable name to identify pointer [OK]

Common Mistakes:

• Thinking * means multiplication here
• Confusing pointer with normal variable
• Ignoring the type before *

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

easy

A. ptr float*;

B. float ptr*;

C. *float ptr;

D. float *ptr;

Solution

1. Step 1: Recall pointer declaration syntax

   Pointer declaration syntax is: type *name; so the * goes between type and variable name.

2. Step 2: Check each option

   float *ptr; matches correct syntax: float *ptr; others have wrong order or missing parts.

3. Final Answer:

   float *ptr; -> Option D

4. Quick Check:

   Pointer syntax = type *name; [OK]

Hint: Always put * between type and variable name [OK]

Common Mistakes:

• Placing * after variable name
• Putting * before type
• Missing * entirely

3. What is the output of this C code?

int x = 10;
int *p = &x;
printf("%d", *p);

medium

A. 10

B. Compilation error

C. Garbage value

D. Address of x

Solution

1. Step 1: Understand pointer initialization

   p is assigned the address of x using &x, so p points to x.

2. Step 2: Dereference pointer to get value

   *p means value at address p, which is x's value 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   *p gives value at pointer address [OK]

Hint: Use * to get value from pointer address [OK]

Common Mistakes:

• Printing pointer p instead of *p
• Confusing address and value
• Forgetting & to get address

4. Identify the error in this pointer declaration:

int *p;
p = 5;

medium

A. Error: assigning integer directly to pointer

B. Correct: p points to integer 5

C. Error: missing * in assignment

D. Error: pointer must be initialized with & operator

Solution

1. Step 1: Check pointer assignment rules

   A pointer must hold a memory address, not a direct integer value.

2. Step 2: Analyze given assignment

   Assigning 5 directly to pointer p is invalid; it should be an address like &variable.

3. Final Answer:

   Error: assigning integer directly to pointer -> Option A

4. Quick Check:

   Pointer = address, not direct value [OK]

Hint: Pointers store addresses, never assign raw values directly [OK]

Common Mistakes:

• Assigning integer instead of address
• Forgetting & operator for address
• Confusing pointer and normal variable

5. Given the code below, which pointer declaration correctly allows modifying the value of num through the pointer?

int num = 20;
// Which declaration is correct?

hard

A. int *ptr = num;

B. int ptr = #

C. int *ptr = #

D. int &ptr = num;

Solution

1. Step 1: Understand pointer declaration for modifying variable

   To modify num via pointer, pointer must hold address of num, declared as int *ptr.

2. Step 2: Check each option

   int *ptr = # correctly declares pointer to int and assigns address of num. Others are invalid syntax or wrong types.

3. Final Answer:

   int *ptr = # -> Option C

4. Quick Check:

   Pointer to int = &variable for modification [OK]

Hint: Use int *ptr = &var to modify var via pointer [OK]

Common Mistakes:

• Assigning value instead of address
• Using wrong type for pointer
• Confusing references with pointers