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C++ Pointers and References

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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.

Opgave

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.

Løsning

solution.cpp

solution.cpp

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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.

Opgave

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.

Løsning

solution.cpp

solution.cpp

Switch to desktopSkift til skrivebord for at øve i den virkelige verdenFortsæt der, hvor du er, med en af nedenstående muligheder
Var alt klart?

Hvordan kan vi forbedre det?

Tak for dine kommentarer!

close

Awesome!

Completion rate improved to 5.88

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