C++ syntax (Java equivalent)

---

Class vs Structs

The only difference between a class and a struct in C++ is that structs have default public members and bases and classes have default private members and bases. Both classes and structs can have a mixture of public, protected and private members, can use inheritance, and can have member functions.

I would recommend using structs as plain-old-data structures without any class-like features, and using classes as aggregate data structures with private data and member functions.

// example about structures
#include <iostream>
#include <string>
#include <sstream>
using namespace std;

struct movies_t {
  string title;
  int year;
} mine, yours;

void printmovie (movies_t movie);

int main ()
{
  string mystr;

  mine.title = "2001 A Space Odyssey";
  mine.year = 1968;

  cout << "Enter title: ";
  getline (cin,yours.title);
  cout << "Enter year: ";
  getline (cin,mystr);
  stringstream(mystr) >> yours.year;

  cout << "My favorite movie is:\n ";
  printmovie (mine);
  cout << "And yours is:\n ";
  printmovie (yours);
  return 0;
}

http://www.cplusplus.com/doc/tutorial/structures/

Parameters passing

1. Copy-constructed objects (pass-by-value):

void cp(point p);

This is pass-by-value. A temporary point object is created and copy-constructed from whatever point object you pass into cp. Once execution leaves the cp function, the temporary object is destroyed.

Note that because the function operates on a copy of whatever was passed to the function, you can modify that local point object as much as you want, the original object of the caller will not be affected. There's no way to change this behaviour with pass-by-value parameters.

2. References to objects (pass-by-reference):

void cp(point& p);

This is pass-by-reference. The actual object passed to the function is available (and potentially subject to modification) inside the function. If you don't want the function to be able to change the passed-in object, declare the parameter as const point&:

void cp(const point& p);

3. Pointers to objects (pass-by-reference):

void cp(point* p);

This is also pass-by-reference, with a few subtle differences. One notable difference is that you can pass a null pointer to the function. This is not possible with references, because references must be initialized and cannot be reseated afterwards. -- As with references, if you want to disallow cp from changing the passed-in point, declare it as const:

void cp(const point* p);

https://stackoverflow.com/questions/2724708/is-it-a-good-practice-to-pass-struct-object-as-parameter-to-a-function-in-c

What is keyword "explicit operator"?

struct A
{
    A(int) { }      // converting constructor
    A(int, int) { } // converting constructor (C++11)
    operator bool() const { return true; }
};
 
struct B
{
    explicit B(int) { }
    explicit B(int, int) { }
    explicit operator bool() const { return true; }
};
 
int main()
{
    A a1 = 1;      // OK: copy-initialization selects A::A(int)
    A a2(2);       // OK: direct-initialization selects A::A(int)
    A a3 {4, 5};   // OK: direct-list-initialization selects A::A(int, int)
    A a4 = {4, 5}; // OK: copy-list-initialization selects A::A(int, int)
    A a5 = (A)1;   // OK: explicit cast performs static_cast
    if (a1) ;      // OK: A::operator bool()
    bool na1 = a1; // OK: copy-initialization selects A::operator bool()
    bool na2 = static_cast<bool>(a1); // OK: static_cast performs direct-initialization
 
//  B b1 = 1;      // error: copy-initialization does not consider B::B(int)
    B b2(2);       // OK: direct-initialization selects B::B(int)
    B b3 {4, 5};   // OK: direct-list-initialization selects B::B(int, int)
//  B b4 = {4, 5}; // error: copy-list-initialization does not consider B::B(int,int)
    B b5 = (B)1;   // OK: explicit cast performs static_cast
    if (b2) ;      // OK: B::operator bool()
//  bool nb1 = b2; // error: copy-initialization does not consider B::operator bool()
    bool nb2 = static_cast<bool>(b2); // OK: static_cast performs direct-initialization
}

https://en.cppreference.com/w/cpp/language/explicit

What is "friendly operator"?

Declaring the function to be a friend of the class, gives it access to protected and private members of the class without opening up access to anyone else.

Overloading the arithmetic operators using friend functions

Some of the most commonly used operators in C++ are the arithmetic operators -- that is, the plus operator (+), minus operator (-), multiplication operator (*), and division operator (/). Note that all of the arithmetic operators are binary operators -- meaning they take two operands -- one on each side of the operator. All four of these operators are overloaded in the exact same way.

It turns out that there are three different ways to overload operators: the member function way, the friend function way, and the normal function way.

#include <iostream>
 
class Cents
{
private:
	int m_cents;
 
public:
	Cents(int cents) { m_cents = cents; }
 
	// add Cents + Cents using a friend function
	friend Cents operator+(const Cents &c1, const Cents &c2);
 
	int getCents() const { return m_cents; }
};
 
// note: this function is not a member function!
Cents operator+(const Cents &c1, const Cents &c2)
{
	// use the Cents constructor and operator+(int, int)
	// we can access m_cents directly because this is a friend function
	return Cents(c1.m_cents + c2.m_cents);
}
 
int main()
{
	Cents cents1(6);
	Cents cents2(8);
	Cents centsSum = cents1 + cents2;
	std::cout << "I have " << centsSum.getCents() << " cents." << std::endl;
 
	return 0;
}

https://www.learncpp.com/cpp-tutorial/92-overloading-the-arithmetic-operators-using-friend-functions/

What is inline function?

C++ inline function is powerful concept that is commonly used with classes. If a function is inline, the compiler places a copy of the code of that function at each point where the function is called at compile time.

Any change to an inline function could require all clients of the function to be recompiled because compiler would need to replace all the code once again otherwise it will continue with old functionality.

What is const function?

• function1 returns a const string pointer

const string *function1()

mean string returns cannot be modified.

• function2 is a const function

char *function2 const()

means that it cannot modify the members of the class, i.e. readonly function.

• function3 params is a const

char function(const data)

mean parameter data cannot be modified.

What is const variable?

#include <array>
const array<int, 3> =  myarray {10,20,30};

means data in array cannot be modified.

http://duramecho.com/ComputerInformation/WhyHowCppConst.html

std::is_base_of

Return true if Derived is derived from Base or if both are the same non-union class (in both cases ignoring cv-qualification)

#include <iostream>
#include <type_traits>
 
class A {};
 
class B : A {};
 
class C {};
 
int main() 
{
    std::cout << std::boolalpha;
    std::cout << "a2b: " << std::is_base_of<A, B>::value << '\n';
    std::cout << "b2a: " << std::is_base_of<B, A>::value << '\n';
    std::cout << "c2b: " << std::is_base_of<C, B>::value << '\n';
    std::cout << "same type: " << std::is_base_of<C, C>::value << '\n';
}

output

a2b: true
b2a: false
c2b: false
same type: true

https://en.cppreference.com/w/cpp/types/is_base_of

std::enable_if

template< bool B, class T = void > struct enable_if;

https://en.cppreference.com/w/cpp/types/enable_if

When to use

• make_delegate?? /// Create a delegate (bound member function)