Meaning of 'Const' Last in a Function Declaration of a Class

What is meant with const at end of function declaration? [duplicate]

A "const function", denoted with the keyword const after a function declaration, makes it a compiler error for this class function to change a member variable of the class. However, reading of a class variables is okay inside of the function, but writing inside of this function will generate a compiler error.

Another way of thinking about such "const function" is by viewing a class function as a normal function taking an implicit this pointer. So a method int Foo::Bar(int random_arg) (without the const at the end) results in a function like int Foo_Bar(Foo* this, int random_arg), and a call such as Foo f; f.Bar(4) will internally correspond to something like Foo f; Foo_Bar(&f, 4). Now adding the const at the end (int Foo::Bar(int random_arg) const) can then be understood as a declaration with a const this pointer: int Foo_Bar(const Foo* this, int random_arg). Since the type of this in such case is const, no modifications of member variables are possible.

It is possible to loosen the "const function" restriction of not allowing the function to write to any variable of a class. To allow some of the variables to be writable even when the function is marked as a "const function", these class variables are marked with the keyword mutable. Thus, if a class variable is marked as mutable, and a "const function" writes to this variable then the code will compile cleanly and the variable is possible to change. (C++11)

As usual when dealing with the const keyword, changing the location of the const key word in a C++ statement has entirely different meanings. The above usage of const only applies when adding const to the end of the function declaration after the parenthesis.

const is a highly overused qualifier in C++: the syntax and ordering is often not straightforward in combination with pointers. Some readings about const correctness and the const keyword:

Const correctness

The C++ 'const' Declaration: Why & How

What is the meaning of a const at end of a member function? [duplicate]

It means that *this is const inside that member function, i.e. it doesn't alter the object.

The keyword this is a prvalue expression whose value is the address of the object for which the function is called. The type of this in a member function of a class X is X*. If the member function is declared const, the type of this is const X*. [section 9.3.2 §1]

In a const member function, the object for which the function is called is accessed through a const access path; therefore, a const member function shall not modify the object and its non-static data members. [section 9.3.2 §2]

This means that a const member function can be called on a const instance of the class. A non-const member function can't be called on ^[1]a const object, since it could potentially try to modify it.

^{[1] Note: a temporary is not a const object unless it's of const type.}

When to use const in the beginning and in the end of a method declaration in C++ [duplicate]

Your first function operates on a const this pointer (that is; a const object that it can't change (or at least shouldn't)).

Your second function returns a constant integer - which is somewhat nonsensical since you can just assign it to a non-const variable and change it anyway. Besides, why does the function care if you change a POD type or not?

Your third function is just a combination of the first two. A function operating on a const object returning a const value.

Const at the end of function declaration in C++ [duplicate]

It means the method is a "const method" A call to such a method cannot change any of the instance's data (with the exception of mutable data members) and can only call other const methods.

Const methods can be called on const or non-const instances, but non-const methods can only be called on non-const instances.

struct Foo {
  void bar() const {}
  void boo() {}
};

Foo f0;
f0.bar(); // OK
fo.boo(); // OK

const Foo f1;
f1.bar(); // OK
f1.boo(); // Error!

why put a const at the end? [duplicate]

The const keyword signifies that the method isn't going to change the object. Since operator== is for comparison, nothing needs to be changed. Therefore, it is const. It has to be omitted for methods like operator= that DO modify the object.

It lets the compiler double-check your work to make sure you're not doing anything you're not supposed to be doing. For more information check out http://www.parashift.com/c++-faq-lite/const-correctness.html.

What does const mean following a function/method signature? [duplicate]

It means that the method do not modify member variables (except for the members declared as mutable), so it can be called on constant instances of the class.

class A
{
public:
    int foo() { return 42; }
    int bar() const { return 42; }
};

void test(const A& a)
{
    // Will fail
    a.foo();

    // Will work
    a.bar();
}

const best practice in function declaration

Assuming that you are talking about member functions (non member functions cannot be const ever), you should write const-correct code always, and that means that each and every function that does not modify the visible state of the object should be const.

Even if you don't ever create a constant object of the type, in many cases you will pass the object to a function through const references and in that case, only const member functions can be called on the object (Incidentally this means that you should pass the object by constant reference to any function that does not need to change the state of the received object)

Why use the keyword 'const' twice in a class member function C++

The first const qualifies the returned value as non-modifiable; the second const specifies that the function does not modify the class instance (or any member of it) on which it was called.

The first is more often used when the function returns a reference to a member, to prevent that returned reference from being used to modify the member to which it refers.

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