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
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 = # 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.
Swipe to start coding
Write a function that swaps the values of two integer variables using pointers.
- Create a function
swapthat 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, callswap, passing the addresses of the two variables.
Løsning
Takk for tilbakemeldingene dine!
single
Spør AI
Spør AI
Spør om hva du vil, eller prøv ett av de foreslåtte spørsmålene for å starte chatten vår
Can you explain how to use the address-of operator when passing a variable?
What is the difference between passing by value and passing by pointer?
Can you give an example of passing a variable by pointer?
Awesome!
Completion rate improved to 5.88
Pointers Use Cases
Sveip for å vise menyen
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 = # 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.
Swipe to start coding
Write a function that swaps the values of two integer variables using pointers.
- Create a function
swapthat 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, callswap, passing the addresses of the two variables.
Løsning
Takk for tilbakemeldingene dine!
single