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C++ Pointers and References

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.

main.cpp

main.cpp

copy

              
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#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.

Tarea

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.

Solución

solution.cpp

solution.cpp

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Sección 1. Capítulo 4
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bookPointers Use Cases

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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.

Tarea

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.

Solución

solution.cpp

solution.cpp

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¿Todo estuvo claro?

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¡Gracias por tus comentarios!

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Completion rate improved to 5.88
Sección 1. Capítulo 4
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