We make a non-member function a friend of a class when we want it to access that class\'s private members. This gives it the same access rights as a static member function w
The standard requires that operator = () [] and -> must be members, and class-specific
operators new, new[], delete and delete[] must be static members. If the situation
arises where we don't need the object of the class to invoke a function, then make
the function static. For all other functions:
if a function requires the operators = () [] and -> for stream I/O,
or if it needs type conversions on its leftmost argument,
or if it can be implemented using the class' public interface alone,
make it nonmember ( and friend if needed in the first two cases)
if it needs to behave virtually,
add a virtual member function to provide the virtual behaviour
and implement in terms of that
else
make it a member.
Here is what I think it is:
Friend function- when you need access to a different class member, but the classes are not related. Static function- when you no not need access to the 'this' pointer. But, I have a feeling there is more to it....
The question seems to address the situation where the programmer needs to introduce a function that does not work on any instance of a class (hence the possibility of choosing a static
member function). Therefore, I will limit this answer to the following design situation, where the choice is between a static function f()
and a friend free function f()
:
struct A
{
static void f(); // Better this...
private:
friend void f(); // ...or this?
static int x;
};
int A::x = 0;
void A::f() // Defines static function
{
cout << x;
}
void f() // Defines friend free function
{
cout << A::x;
}
int main()
{
A::f(); // Invokes static function
f(); // Invokes friend free function
}
Without knowing anything in advance about the semantics of f()
and A
(I'll come back to this later), this limited scenario has an easy answer: the static
function is preferable. I see two reasons for this.
GENERIC ALGORITHMS:
The main reason is that a template such as the following can be written:
template<typename T> void g() { T::f(); }
If we had two or more classes that have a static
function f()
on their interface, this would allow us writing one single function that invokes f()
generically on any such class.
There is no way to write an equivalent generic function if we make f()
a free, non-member function. Although it is true that we could put f()
into a namespace, so that the N::f()
syntax could be used to mimic the A::f()
syntax, it would still be impossible to write a template function such as g<>()
above, because namespace names are not valid template arguments.
REDUNDANT DECLARATIONS:
The second reason is that if we were to put the free function f()
in a namespace, we would not be allowed to inline its definition directly in the class definition without introducing any other declaration for f()
:
struct A
{
static void f() { cout << x; } // OK
private:
friend void N::f() { cout << x; } // ERROR
static int x;
};
In order to fix the above, we would to preceed the definition of class A
with the following declaration:
namespace N
{
void f(); // Declaration of f() inside namespace N
}
struct A
{
...
private:
friend void N::f() { cout << x; } // OK
...
};
This, however, defeats our intention of having f()
declared and defined in just one place.
Moreover, if we wanted to declare and define f()
separately while keeping f()
in a namespace, we would still have to introduce a declaration for f()
before the class definition for A
: failing to do so would cause the compiler to complain about the fact that f()
had to be declared inside namespace N
before the name N::f
could be used legally.
Thus, we would now have f()
mentioned in three separate places rather than just two (declaration and definition):
N
before A
's definition;friend
declaration inside A
's definition;f()
inside namespace N
.The reason why the declaration and definition of f()
inside N
cannot be joined (in general) is that f()
is supposed to access the internals of A
and, therefore, A
's definition must be seen when f()
is defined. Yet, as previously said, f()
's declaration inside N
must be seen before the corresponding friend
declaration inside of A
is made. This effectively forces us to split the declaration and the definition of f()
.
SEMANTIC CONSIDERATIONS:
While the above two points are universally valid, there are reasons why one might prefer declaring f()
as static
over making it a friend
of A
or vice versa which are driven by the universe of discourse.
To clarify, it is important to stress the fact that a member function of a class, whether it is static
or non-static
, is logically part of that class. It contributes to its definition and thus provides a conceptual characterization of it.
On the other hand, a friend
function, in spite of being granted access to the internal members of the class it is friend of, is still an algorithm which is logically external to the definition of the class.
A function can be friend
of more than one class, but it can be member of just one.
Thus, in a particular application domain, the designer may want to keep into consideration the semantics of both the function and the class when deciding whether to make the former a friend
or a member of the latter (this applies not only to static
functions, but to non-static
functions as well, where other language constraints may intervene).
Does the function logically contribute to characterize a class and/or its behavior, or is it rather an external algorithm? This question can't be answered without knowledge of the particular application domain.
TASTE:
I believe that any argument other the ones just given stems purely from a matter of taste: both the free friend
and the static
member approach, in fact, allow to clearly state what the interface of a class is into one single spot (the class's definition), so design-wise they are equivalent (modulo the above observations, of course).
The remaining differences are stylistic: whether we want to write the static
keyword or the friend
keyword when declaring a function, and whether we want to write the A::
class scope qualifier when defining the class rather than the N::
namespace scope qualifier. Thus, I will not comment further on this.