If a member function is defined inside the class, it is an inline function. E.g.
struct X
{
void mem_f() {} //mem_f is inline
};
My question is whether a nonmember friend function defined inside the class is also automatically inline.
E.g.
struct Y
{
friend void friend_f() {} //is friend_f inline?
};
A relevant quote/paragraph_no from the standard would be much welcome. Thanks.
Yes, it is. §11.4/5:
A function can be defined in a friend
declaration of a class if and only if
the class is a non-local class (9.8),
the function name is unqualified, and
the function has namespace scope.
Such a function is implicitly inline. A friend function defined in
a class is in the (lexical) scope of
the class in which it is defined. A
friend function defined outside the
class is not (3.4.1).
Since the class definition is presumably in a header file, the function will be multiply-defined, so it needs to be inline.
Related
Can the scope resolution operator be used in the definition of a member function when the member function is defined outside of its class definition? Like this:
class Myclass{
private:
void foo();
public:
void foo2();
};
void Myclass::foo()
{
// do stuff. . .
}
My teacher gave us notes that demonstrate the use of the scope resolution operator only for member functions that are public. Can anyone clear this up?
Why doesn't this compile?
class Test
{
void foo();
void foo()
{ }
};
But these will compile:
void bar();
void bar()
{ }
// In same header/file
class Test
{
void foo();
};
void Test::foo()
{ }
The compiler would say that given method cannot be overloaded. The method Test::foo is not being overloaded - it is the same function with exact same signature.
It's explicitly prohibited by the C++ standard. Right there at [class.mfct/1]
A member function may be defined in its class definition, in which
case it is an inline member function, or it may be defined outside of
its class definition if it has already been declared but not defined
in its class definition. A member function definition that appears
outside of the class definition shall appear in a namespace scope
enclosing the class definition. Except for member function definitions
that appear outside of a class definition, and except for explicit
specializations of member functions of class templates and member
function templates ([temp.spec]) appearing outside of the class
definition, a member function shall not be redeclared.
It is just the way it is in C++. Redeclaring a class member is not allowed (unless you consider an out-of-class definition as another declaration). Redeclaring a namespace member is OK as long as you obey ODR.
Why doesn't this compile?
class Test {
void foo();
void foo() { }; // wrong
};
Because, as StoryTeller and others answered, you are declaring and defining the same member function void Test::foo(void) twice (and the inside-class definition of foo is implicitly inline).
If you want to define a member function in your header file after the class, you'll better declare it explicitly as inline like this:
class Test {
inline void foo();
};
and later define that member function (e.g. below in the same header file):
void Test::foo() {
// body of Test::foo
}
BTW, if you declare a member function with outside definition like above, and that member function is not inline but is defined in some header file which is included in several translation units, that function would be multiply defined and the linker would complain.
void foo();
and
void foo() {}
are two ways of declaring functions in a class definition. The first only declares the function and the second declares the function and also implements it. Therefore the compiler assumes that you are going to re-declare the same function and is not a correct overload as the function signature is the same in both.
I am very new to C++, and when I am trying to learn the friend function, I saw from friend description on Cppreference that:
2) (only allowed in non-local class definitions) Defines a non-member function, and makes it a friend of this class at the same time. Such non-member function is always inline.
class X {
int a;
friend void friend_set(X& p, int i) {
p.a = i; // this is a non-member function
}
public:
void member_set(int i) {
a = i; // this is a member function
}
};
Does this mean that all friend functions always have to be inline? In another word, must the friend functions be defined completely inside the class?
However, I also found an instance where the friend function is defined outside the class from Cplusplus
// friend functions
#include <iostream>
using namespace std;
class Rectangle {
int width, height;
public:
Rectangle() {}
Rectangle (int x, int y) : width(x), height(y) {}
int area() {return width * height;}
friend Rectangle duplicate (const Rectangle&);
};
Rectangle duplicate (const Rectangle& param)
{
Rectangle res;
res.width = param.width*2;
res.height = param.height*2;
return res;
}
I am really confused by this conflict. Is my understanding of inline wrong?
What does it mean by "a nonmember friend function defined inside the class is automatically inline"?
any function defined inside a class (member or non-member friend) is always implicitly inline. That's because class definitions are generally in header files, and you don't want any non-inline function definitions in a header file (having a non-inline function in a header leads to multiple definitions if the header is #included in more than one source file).
If you want a function to be non-inline, you need to define it outside of a class definition. If its a member or friend you still need to declare it inside the class, since members and friends must be declared in the class. You just don't want it defined in the class.
This all gets into the difference between a definition and a declaration in C++ -- they are two distinct things, though every definition is also implicitly a declaration, not every declaration is a definition. As a result, when the spec says 'declaration' it often means 'declaration that is not also a definition'
what's wrong with this code?
class matrix{
private:
friend transpose(const matrix&);
friend class invert;
public: //...
};
matrix (*p)(const matrix&)=&transpose; //error: no transpose() in scope.
what does the statement means "a friend declaration does not introduce a name into enclosing scope".This problem does not occur when friend keyword is removed
The difference between the declaration of transpose() as a friend and without the friend declaration is that if you declare "friend transpose()" all you are doing is telling the compiler that a function friend with the signature shown in the friend declaration can have access to the private members of an object of type matrix. It does not declare a function transpose() with this signature - you still have to do this outside the scope of the matrix class.
If you remove the 'friend' keyword, you are declaring a member function transpose() inside the class matrix, so the compiler actually has seen a function it can take the address of.
§7.3.1.2 [namespace.memdef] p3
[...] If a friend declaration in a nonlocal class first declares a class or function the friend class or function is a member of the innermost enclosing namespace. The name of the friend is not found by unqualified lookup or by qualified lookup until a matching declaration is provided in that namespace scope (either before or after the class definition granting friendship). [...]
See also this question of mine.
Friend functions are functions that are not members of a class but
they still have access to the private members of the class.
I should point out that a friend function declaration
may be placed in either the private section or the public section,
but it will be a public function in either case, so it is clearer to
list it in the public section.
class MyClass
{
private:
int data;
public:
MyClass(int value);
friend void myFriend(MyClass *myObject);
};
void myFriend(MyClass *myObject)
{
cout << myObject->data<< endl;
}
MyClass::MyClass(int value)
{
data = value*2;
}
int main()
{
MyClass myObject(3);
myFriend(&myObject);
}
So, you need to define the friend function after you declare it.
After searching aroung SO, one question taught me that the lexical scope of an inline friend function is the class it's defined in, meaning it can access e.g. the typedefs in the class without qualifying them. But then I wondered what is the actual scope of such a function? GCC at least rejects all my attempts to call it. Can a function such as in the example ever be called through means other than ADL, which is not possible here thanks to no arguments?
Standard quotations are appreciated, as I currently can't access my copy of it.
The following code
namespace foo{
struct bar{
friend void baz(){}
void call_friend();
};
}
int main(){
foo::baz(); // can't access through enclosing scope of the class
foo::bar::baz(); // can't access through class scope
}
namespace foo{
void bar::call_friend(){
baz(); // can't access through member function
}
}
results in these errors:
prog.cpp: In function ‘int main()’:
prog.cpp:9: error: ‘baz’ is not a member of ‘foo’
prog.cpp:10: error: ‘baz’ is not a member of ‘foo::bar’
prog.cpp: In member function ‘void foo::bar::call_friend()’:
prog.cpp:15: error: ‘baz’ was not declared in this scope
When you declare a friend function with an unqualified id in a class it names a function in the nearest enclosing namespace scope.
If that function hasn't previously been declared then the friend declaration doesn't make that function visible in that scope for normal lookup. It does make the declared function visible to argument-dependent lookup.
This is emphasised in many notes, but the definitive statement is in 7.3.1.2/3 (of ISO/IEC 14882:2011):
Every name first declared in a namespace is a member of that namespace. If a friend declaration in a non-local class first declares a class or function the friend class or function is a member of the innermost enclosing namespace. The name of the friend is not found by unqualified lookup (3.4.1) or by qualified lookup (3.4.3) until a matching declaration is provided in that namespace scope (either before or after the class definition granting friendship). If a friend function is called, its name may be found by the name lookup that considers functions from namespaces and classes associated with the types of the function arguments (3.4.2). If the name in a friend declaration is neither qualified nor a template-id and the declaration is a function or an elaborated-type-specifier, the lookup to determine whether the entity has been previously declared shall not consider any scopes outside the innermost enclosing namespace.
"The C++ Programming Language 3rd Edition (Stroustrap)" : p279:
I. "Like a member declaration, a friend declaration does not introduce a name into an enclosing scope"
II. "A friend class must be previously declared in an enclosing scope or defined in the nonclass
scope immediately enclosing the class that is declaring it a friend"
III. "A friend function can be explicitly declared just like friend classes, or it can be found through its argument types (§8.2.6) as if it was declared in the nonclass
scope immediately enclosing its class."
IV. "It follows that a friend function should either be explicitly declared in an enclosing scope or take an argument of its class. If not, the friend cannot be called. For example:"
//no f() here
void g();
class X{
friend void f(); //useless
friend void g(); //can be found because it is declared outside of class scope
friend void h(const X&); //can be found because the arguments access class members
};
void f() { } //enemy of X :)
But in your case there is more to it that has to do with the namespace, because if you put the proper declaration in foo e.g.:
namespace foo{
struct bar{
friend void baz(const &bar){};
void call_friend();
}
}
does not compile. However, if you declare it outside foo is works like a charm. Now consider that in fact, global, local, struct, and classes are in fact namespaces. Now this leads to the conclusion that the baz(const &) is implicitly defined in the global scope.
This compiles:
namespace foo{
struct bar{
friend void baz(const bar&){};
void call_friend();
};
}
int main(){
foo::bar k;
baz(k);
return 0;
}
Therefore, there are two issues:
The friend declaration does not introduce a name in an enclosing scope, unless IV. Thus the original program cannot find baz() because it has not been properly declared.
If IV , i.e. ADL, then the function is found in foo, but cannot be accessed as foo::baz(k), due to ADL. You will have to explicitely define baz(const bar&) in foo to access it by qualified name.
Thanks, hope it helps, but certainly, I liked the challenge :) .
Interesting!
It seems that the compiler does not know what scope it belongs to (and to be honest there are no clues) and thus puts in in no scope. Some standard digging coming up I suppose.
Note: If you explicitly add a declaration to a particular scope then it starts to work as expected.
namespace foo
{
void baz(); // declare it here and now it works in foo namespace etc.
struct bar
{
friend void baz(){}
void call_friend();
};
}
Digging the standard I find:
11.3 Friends [class.friend]
Paragraph 6
A function can be defined in a friend declaration of a class if and only if the class is a non-local class (9.8), the function name is unqualified, and the function has namespace scope.
[ Example:
class M { friend void f() { } // definition of global f, a friend of M,
// not the definition of a member function
};
— end example ]
Paragraph 7
Such a function is implicitly inline. A friend function defined in a class is in the (lexical) scope of the class in which it is defined. A friend function defined outside the class is not (3.4.1).
Note:
A free standing function that does not take a parameter is not much use as a friend. As it will have no object on which to take advantage of its friendship (I suppose file scope static storage duration objects).
A friend function defined within an enclosing class, will only be found by argument dependent lookup (ADL). Calling it will succeed when one or more of the arguments is either of the enclosing class type; or of a type declared within the class. Here is an example, displaying "Hello World!", which (for variety) doesn't use an object of the enclosing class type to provide such an argument:
#include <iostream>
struct foo
{
struct local_class{};
friend void greet(local_class o, int) { std::cout << "Hello World!\n"; }
};
int main(int argc, char *argv[])
{
foo::local_class o;
greet(o,1);
return 0;
}
In this Example,
namespace foo{
struct bar{
friend void baz(){}
void call_friend();
};
}
int main(){
foo::baz(); // can't access through enclosing scope of the class
foo::bar::baz(); // can't access through class scope
}
namespace foo{
void bar::call_friend(){
baz(); // can't access through member function
}
}
foo::baz() is inaccessible because the name baz is not visible in scope of namespace foo. If I remember correctly (§ 3.4/2) applies here.
foo::bar::baz() is inaccessible because friends aren't members of class and the scope of inline friend function is namespace or class in which their definition exists therefore, you can't access them outside of that scope.
If you put the declaration of baz() in foo the name of function baz will be visible in foo and the definition will be looked up in the nested scope of bar.
namespace foo{
void baz(); // declaration at namespace scope
struct bar{
friend void baz(){}
};
void call_friend() {
baz(); // ok
}
}
int main()
{
foo::baz(); // ok, bar will be looked up in the nested scope of foo::bar.
}
I think you are confusing friend and private. By declaring function a friend you are granting it access to your private members, and not grating other functions access to it. Anyway, any member function of a struct is accessible by any object because struct members are public by default.
However, in your case baz isn't accessible because by doing friend void baz(){} you didn't really declare the function baz, you just said that it is a friend function. You can just remove the friend keyword, and it will solve all the issues.