Lernen Verwendungsfälle von Zeigern

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


              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


              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.

Aufgabe

Swipe to start coding

Write a function that swaps the values of two integer variables using pointers.

Create a function swap that takes two pointers to integers as parameters.
Inside the function, use a temporary variable to store the value of the first variable.
Assign the value of the second variable to the first variable using pointer dereferencing.
Assign the value from the temporary variable to the second variable using pointer dereferencing.
In main, call swap, passing the addresses of the two variables.

Lösung

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Abschnitt 1. Kapitel 4

single

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main.cpp


              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


              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.

Aufgabe

Swipe to start coding

Write a function that swaps the values of two integer variables using pointers.

Create a function swap that takes two pointers to integers as parameters.
Inside the function, use a temporary variable to store the value of the first variable.
Assign the value of the second variable to the first variable using pointer dereferencing.
Assign the value from the temporary variable to the second variable using pointer dereferencing.
In main, call swap, passing the addresses of the two variables.

Lösung

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War alles klar?

Danke für Ihr Feedback!

Abschnitt 1. Kapitel 4

single