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Pointers Use Cases | Pointers Fundamentals
C++ Pointers and References
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Зміст курсу

C++ Pointers and References

C++ Pointers and References

1. Pointers Fundamentals
2. Pointer Arithmetic
3. References Fundamentals
4. Dynamic Memory Allocation

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.

cpp

main

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.

cpp

main

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.

Завдання

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

Завдання

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

Все було зрозуміло?

Секція 1. Розділ 4
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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.

cpp

main

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.

cpp

main

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.

Завдання

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

Завдання

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

Все було зрозуміло?

Секція 1. Розділ 4
toggle bottom row

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.

cpp

main

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.

cpp

main

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.

Завдання

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

Завдання

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

Все було зрозуміло?

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.

cpp

main

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.

cpp

main

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.

Завдання

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

Секція 1. Розділ 4
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