C++ – Override and overload in C++

c++overloadingoverriding

Yes, I do understand the difference between them. What I want to know is: why OVERRIDE a method? What is the good in doing it?
In case of overload: the only advantage is you haven't to think in different names to functions?

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Overloading generally means that you have two or more functions in the same scope having the same name. The function that better matches the arguments when a call is made wins and is called. Important to note, as opposed to calling a virtual function, is that the function that's called is selected at compile time. It all depends on the static type of the argument. If you have an overload for B and one for D, and the argument is a reference to B, but it really points to a D object, then the overload for B is chosen in C++. That's called static dispatch as opposed to dynamic dispatch. You overload if you want to do the same as another function having the same name, but you want to do that for another argument type. Example:

void print(Foo const& f) {
    // print a foo
}

void print(Bar const& bar) {
    // print a bar
}

they both print their argument, so they are overloaded. But the first prints a foo, and the second prints a bar. If you have two functions that do different things, it's considered bad style when they have the same name, because that can lead to confusion about what will happen actually when calling the functions. Another usecase for overloading is when you have additional parameters for functions, but they just forward control to other functions:

void print(Foo & f, PrintAttributes b) { 
    /* ... */ 
}

void print(Foo & f, std::string const& header, bool printBold) {
    print(f, PrintAttributes(header, printBold));
}

That can be convenient for the caller, if the options that the overloads take are often used.

Overriding is something completely different. It doesn't compete with overloading. It means that if you have a virtual function in a base class, you can write a function with the same signature in the derived class. The function in the derived class overrides the function of the base. Sample:

struct base {
    virtual void print() { cout << "base!"; }
}

struct derived: base {
    virtual void print() { cout << "derived!"; }
}

Now, if you have an object and call the print member function, the print function of the derived is always called, because it overrides the one of the base. If the function print wasn't virtual, then the function in the derived wouldn't override the base function, but would merely hide it. Overriding can be useful if you have a function that accepts a base class, and every one that's derived from it:

void doit(base &b) {
    // and sometimes, we want to print it
    b.print();
}

Now, even though at compile time the compiler only knows that b is at least base, print of the derived class will be called. That's the point of virtual functions. Without them, the print function of the base would be called, and the one in the derived class wouldn't override it.

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C++ – What are the differences between a pointer variable and a reference variable in C++

A pointer can be re-assigned:

int x = 5;
int y = 6;
int *p;
p = &x;
p = &y;
*p = 10;
assert(x == 5);
assert(y == 10);

A reference cannot be re-bound, and must be bound at initialization:

int x = 5;
int y = 6;
int &q; // error
int &r = x;

A pointer variable has its own identity: a distinct, visible memory address that can be taken with the unary & operator and a certain amount of space that can be measured with the sizeof operator. Using those operators on a reference returns a value corresponding to whatever the reference is bound to; the reference’s own address and size are invisible. Since the reference assumes the identity of the original variable in this way, it is convenient to think of a reference as another name for the same variable.
```
int x = 0;
int &r = x;
int *p = &x;
int *p2 = &r;

assert(p == p2); // &x == &r
assert(&p != &p2);
```

You can have arbitrarily nested pointers to pointers offering extra levels of indirection. References only offer one level of indirection.

int x = 0;
int y = 0;
int *p = &x;
int *q = &y;
int **pp = &p;

**pp = 2;
pp = &q; // *pp is now q
**pp = 4;

assert(y == 4);
assert(x == 2);

A pointer can be assigned nullptr, whereas a reference must be bound to an existing object. If you try hard enough, you can bind a reference to nullptr, but this is undefined and will not behave consistently.

/* the code below is undefined; your compiler may optimise it
 * differently, emit warnings, or outright refuse to compile it */

int &r = *static_cast<int *>(nullptr);

// prints "null" under GCC 10
std::cout
    << (&r != nullptr
        ? "not null" : "null")
    << std::endl;

bool f(int &r) { return &r != nullptr; }

// prints "not null" under GCC 10
std::cout
    << (f(*static_cast<int *>(nullptr))
        ? "not null" : "null")
    << std::endl;

You can, however, have a reference to a pointer whose value is nullptr.

Pointers can iterate over an array; you can use ++ to go to the next item that a pointer is pointing to, and + 4 to go to the 5th element. This is no matter what size the object is that the pointer points to.
A pointer needs to be dereferenced with * to access the memory location it points to, whereas a reference can be used directly. A pointer to a class/struct uses -> to access its members whereas a reference uses a ..
References cannot be put into an array, whereas pointers can be (Mentioned by user @litb)
Const references can be bound to temporaries. Pointers cannot (not without some indirection):
```
const int &x = int(12); // legal C++
int *y = &int(12); // illegal to take the address of a temporary.
```
This makes const & more convenient to use in argument lists and so forth.

C++ – What should main() return in C and C++

The return value for main indicates how the program exited. Normal exit is represented by a 0 return value from main. Abnormal exit is signaled by a non-zero return, but there is no standard for how non-zero codes are interpreted. As noted by others, void main() is prohibited by the C++ standard and should not be used. The valid C++ main signatures are:

int main()

and

int main(int argc, char* argv[])

which is equivalent to

int main(int argc, char** argv)

It is also worth noting that in C++, int main() can be left without a return-statement, at which point it defaults to returning 0. This is also true with a C99 program. Whether return 0; should be omitted or not is open to debate. The range of valid C program main signatures is much greater.

Efficiency is not an issue with the main function. It can only be entered and left once (marking the program's start and termination) according to the C++ standard. For C, re-entering main() is allowed, but should be avoided.

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C++ – What are the differences between a pointer variable and a reference variable in C++

C++ – What should main() return in C and C++

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