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C++ Pointers and References
course content

Course Content

C++ Pointers and References

C++ Pointers and References

1. Pointers Fundamentals
Introduction to PointerAssigning Addresses to PointersLink Between Pointer and VariablePointers Use Cases
2. Pointer Arithmetic
Introduction to Pointer ArithmeticPointer Arithmetic PracticeArrays and Pointers ConnectionArray Pointer Arithmetic Practice
3. References Fundamentals
Introduction to ReferencesReference Use CasesRvalue and Lvalue ReferencesReference vs Pointers
4. Dynamic Memory Allocation
Introduction to MemoryDynamic AllocationDynamic Allocation of the ArrayPointer to Pointer Dynamic AllocationSummary and Overview

bookPointers Use Cases

When you pass a variable to a function, you're essentially passing its value. This means the function receives a copy of the data. Any modifications made inside the function DO NOT affect the original variable.

cpp

main

copy

              
12345678910

            
#include <iostream>

void increment(int num) { num++; }

int main() 
{
    int num = 5;
    increment(num);
    std::cout << "Original value: " << num << std::endl;
}

We can use pointers to enable a function to alter the original variable. This involves passing a memory address as an argument instead of the actual value.

cpp

main

copy

              
123456789101112

            
#include <iostream>

void increment(int* num) { (*num)++; }

int main() 
{
	int num = 5;
    int* p_num = &num;	

    increment(p_num);
    std::cout << "Original value: " << num << std::endl;
}

Note

You can bypass the creation of a pointer to a variable and instead directly use the address-of operator when passing a variable.

Task

  • Create a function that swaps values of two variables.
  • Call this function.
  • Output the values of variables after the swap.

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Section 1. Chapter 4
toggle bottom row

bookPointers Use Cases

When you pass a variable to a function, you're essentially passing its value. This means the function receives a copy of the data. Any modifications made inside the function DO NOT affect the original variable.

cpp

main

copy

              
12345678910

            
#include <iostream>

void increment(int num) { num++; }

int main() 
{
    int num = 5;
    increment(num);
    std::cout << "Original value: " << num << std::endl;
}

We can use pointers to enable a function to alter the original variable. This involves passing a memory address as an argument instead of the actual value.

cpp

main

copy

              
123456789101112

            
#include <iostream>

void increment(int* num) { (*num)++; }

int main() 
{
	int num = 5;
    int* p_num = &num;	

    increment(p_num);
    std::cout << "Original value: " << num << std::endl;
}

Note

You can bypass the creation of a pointer to a variable and instead directly use the address-of operator when passing a variable.

Task

  • Create a function that swaps values of two variables.
  • Call this function.
  • Output the values of variables after the swap.

Switch to desktopSwitch to desktop for real-world practiceContinue from where you are using one of the options below
Everything was clear?

How can we improve it?

Thanks for your feedback!

Section 1. Chapter 4
toggle bottom row

bookPointers Use Cases

When you pass a variable to a function, you're essentially passing its value. This means the function receives a copy of the data. Any modifications made inside the function DO NOT affect the original variable.

cpp

main

copy

              
12345678910

            
#include <iostream>

void increment(int num) { num++; }

int main() 
{
    int num = 5;
    increment(num);
    std::cout << "Original value: " << num << std::endl;
}

We can use pointers to enable a function to alter the original variable. This involves passing a memory address as an argument instead of the actual value.

cpp

main

copy

              
123456789101112

            
#include <iostream>

void increment(int* num) { (*num)++; }

int main() 
{
	int num = 5;
    int* p_num = &num;	

    increment(p_num);
    std::cout << "Original value: " << num << std::endl;
}

Note

You can bypass the creation of a pointer to a variable and instead directly use the address-of operator when passing a variable.

Task

  • Create a function that swaps values of two variables.
  • Call this function.
  • Output the values of variables after the swap.

Switch to desktopSwitch to desktop for real-world practiceContinue from where you are using one of the options below
Everything was clear?

How can we improve it?

Thanks for your feedback!

When you pass a variable to a function, you're essentially passing its value. This means the function receives a copy of the data. Any modifications made inside the function DO NOT affect the original variable.

cpp

main

copy

              
12345678910

            
#include <iostream>

void increment(int num) { num++; }

int main() 
{
    int num = 5;
    increment(num);
    std::cout << "Original value: " << num << std::endl;
}

We can use pointers to enable a function to alter the original variable. This involves passing a memory address as an argument instead of the actual value.

cpp

main

copy

              
123456789101112

            
#include <iostream>

void increment(int* num) { (*num)++; }

int main() 
{
	int num = 5;
    int* p_num = &num;	

    increment(p_num);
    std::cout << "Original value: " << num << std::endl;
}

Note

You can bypass the creation of a pointer to a variable and instead directly use the address-of operator when passing a variable.

Task

  • Create a function that swaps values of two variables.
  • Call this function.
  • Output the values of variables after the swap.

Switch to desktopSwitch to desktop for real-world practiceContinue from where you are using one of the options below
Section 1. Chapter 4
Switch to desktopSwitch to desktop for real-world practiceContinue from where you are using one of the options below
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