Notice: This page requires JavaScript to function properly.
Please enable JavaScript in your browser settings or update your browser.Leer Pointers Use Cases | Pointers Fundamentals
C++ Pointers and References

Veeg om het menu te tonen

book
Pointers 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.

main.cpp

main.cpp

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.

main.cpp

main.cpp

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.

Taak

Swipe to start coding

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

Oplossing

solution.cpp

solution.cpp

Switch to desktopSchakel over naar desktop voor praktijkervaringGa verder vanaf waar je bent met een van de onderstaande opties
Was alles duidelijk?

Hoe kunnen we het verbeteren?

Bedankt voor je feedback!

Sectie 1. Hoofdstuk 4
single

single

Vraag AI

expand

Vraag AI

ChatGPT

Vraag wat u wilt of probeer een van de voorgestelde vragen om onze chat te starten.

close

Awesome!

Completion rate improved to 5.88

book
Pointers 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.

main.cpp

main.cpp

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.

main.cpp

main.cpp

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.

Taak

Swipe to start coding

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

Oplossing

solution.cpp

solution.cpp

Switch to desktopSchakel over naar desktop voor praktijkervaringGa verder vanaf waar je bent met een van de onderstaande opties
Was alles duidelijk?

Hoe kunnen we het verbeteren?

Bedankt voor je feedback!

close

Awesome!

Completion rate improved to 5.88

Veeg om het menu te tonen

some-alt