Pointers to pointers in C

Introduction

Pointers to pointers let you store the address of a pointer. This helps when you want to change a pointer inside a function or manage multiple levels of data.

When you want a function to change the address stored in a pointer.

When working with arrays of strings (like list of words).

When managing dynamic multi-dimensional arrays.

When you need to pass a pointer by reference to another function.

Syntax

C

type **pointer_to_pointer;

Each * adds one level of indirection (address pointing).

Use ** to declare a pointer that points to another pointer.

Examples

Declares a pointer to a pointer to an int.

C

int **pptr;

Declares a pointer to a pointer to a char (often used for strings).

C

char **str_ptr;

Pointer pptr stores the address of pointer ptr.

C

int *ptr;
int **pptr = &ptr;

Sample Program

This program shows how a pointer to a pointer can access and change the original value.

C

#include <stdio.h>

int main() {
    int value = 10;
    int *ptr = &value;       // pointer to int
    int **pptr = &ptr;       // pointer to pointer to int

    printf("Value: %d\n", value);
    printf("Value via ptr: %d\n", *ptr);
    printf("Value via pptr: %d\n", **pptr);

    // Change value using pointer to pointer
    **pptr = 20;
    printf("New value after change via pptr: %d\n", value);

    return 0;
}

Important Notes

Remember to use ** to access the value when using a pointer to a pointer.

Pointer to pointer is useful for modifying pointers inside functions.

Summary

Pointer to pointer stores the address of another pointer.

Use ** to declare and access pointer to pointer values.

It helps when you want to change pointers or work with multi-level data.

Practice

1. What does a pointer to a pointer store in C?

easy

A. The size of a variable

B. The value of an integer

C. The address of another pointer

D. The address of a function

Solution

1. Step 1: Understand pointer basics

   A pointer stores the address of a variable.

2. Step 2: Understand pointer to pointer

   A pointer to pointer stores the address of another pointer, not a direct value.

3. Final Answer:

   The address of another pointer -> Option C

4. Quick Check:

   Pointer to pointer = address of pointer [OK]

Hint: Pointer to pointer holds address of pointer, not value [OK]

Common Mistakes:

• Confusing pointer to pointer with pointer to value
• Thinking it stores the value directly
• Mixing it with function pointers

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

easy

A. int **ptr;

B. int ptr*;

C. int *ptr*;

D. int &ptr*;

Solution

1. Step 1: Recall pointer declaration syntax

   Single pointer to int is declared as int *ptr;.

2. Step 2: Extend to pointer to pointer

   Pointer to pointer uses two asterisks: int **ptr;.

3. Final Answer:

   int **ptr; -> Option A

4. Quick Check:

   Pointer to pointer declaration uses two * [OK]

Hint: Use two * for pointer to pointer declaration [OK]

Common Mistakes:

• Placing * incorrectly in declaration
• Using & instead of *
• Confusing pointer syntax with references

3. What is the output of the following C code?

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

medium

A. Address of x

B. Compilation error

C. Address of p

D. 10

Solution

1. Step 1: Understand variable assignments

   x = 10, p points to x, pp points to p.

2. Step 2: Evaluate **pp

   *pp gives p (pointer to x), **pp gives value at x, which is 10.

3. Final Answer:

   10 -> Option D

4. Quick Check:

   **pp dereferences twice to get x's value [OK]

Hint: Double dereference **pp gets original value [OK]

Common Mistakes:

• Printing pointer address instead of value
• Using single * instead of **
• Confusing *p and **pp

4. Identify the error in the following code snippet:

int x = 5;
int *p = &x;
int **pp;
*pp = p;
printf("%d", **pp);

medium

A. Syntax error in printf

B. pp is not initialized before dereferencing

C. x is not initialized

D. p is not assigned correctly

Solution

1. Step 1: Check pointer initialization

   pp is declared but not initialized to point anywhere.

2. Step 2: Analyze dereferencing *pp = p;

   Dereferencing uninitialized pp causes undefined behavior or crash.

3. Final Answer:

   pp is not initialized before dereferencing -> Option B

4. Quick Check:

   Dereferencing uninitialized pointer causes error [OK]

Hint: Always initialize pointers before dereferencing [OK]

Common Mistakes:

• Assuming pointer auto-initializes
• Ignoring uninitialized pointer usage
• Confusing pointer assignment with value assignment

5. Given the code below, what will be the output?

int a = 3, b = 7;
int *p1 = &a;
int *p2 = &b;
int **pp = &p1;
*pp = p2;
printf("%d %d", *p1, *p2);

hard

A. 7 7

B. 3 7

C. 3 3

D. Compilation error

Solution

1. Step 1: Initial pointer assignments

   p1 points to a (3), p2 points to b (7), pp points to p1.

2. Step 2: Change p1 via pointer to pointer

   *pp = p2 changes p1 to point to b (7).

3. Step 3: Print values

   *p1 now points to b (7), *p2 is 7, so output is "7 7".

4. Final Answer:

   7 7 -> Option A

5. Quick Check:

   Pointer to pointer changed p1 to point to b [OK]

Hint: Pointer to pointer can change pointer target [OK]

Common Mistakes:

• Assuming p1 still points to a
• Confusing *pp and **pp
• Ignoring pointer reassignment effect