I wrote this code in C++:
class Foo
{
public:
int& fun(){return var;} // 1st fun
int fun() const {return var;} // 2rd fun
private:
int var;
};
int main()
{
Foo foo;
int i = foo.fun();
return 0;
}
I know that C++ cannot discriminate overloading function by return value,but why when I added a const to 2rd function ,overloading can work ?
What the 'const' have done ?
Compiler cannot discriminate by return type because return values can undergo conversion before the assignment is performed. The object on which the function is invoked, on the other hand, is a parameter (albeit an implicit one) to the function, so the compiler can discriminate on it.
Const is used in the following way:
Foo inst1;
const Foo inst2;
inst1.fun(); // 1st fun
inst2.fun(); // 2nd fun
Const after the name of the function refers to the implicit this parameter. So, for inst1 it will Foo* and for inst2 const Foo*. This will guide the overload.
The return value is not used for selecting the overload. Methods/functions with the same set of params and different types of return value are not allowed on the same layer.
For what it's worth, the language from the standard (§13.3.1/3,4):
Similarly, when appropriate, the context can construct an argument list that contains an implied object
argument to denote the object to be operated on. Since arguments and parameters are associated by position within their respective lists, the convention is that the implicit object parameter, if present, is
always the first parameter and the implied object argument, if present, is always the first argument.
For non-static member functions, the type of the implicit object parameter is
— “lvalue reference to cv X” for functions declared without a ref-qualifier or with the & ref-qualifier
— “rvalue reference to cv X” for functions declared with the && ref-qualifier
where X is the class of which the function is a member and cv is the cv-qualification on the member function declaration. [ Example: for a const member function of class X, the extra parameter is assumed to have type “reference to const X”. —end example ]
Related
In this link : Implicit object parameter
In this quote :
If any candidate function is a member function (static or non-static) that does not have an explicit object parameter (since C++23), but not a constructor, it is treated as if it has an extra parameter (implicit object parameter) which represents the object for which they are called and appears before the first of the actual parameters.
I do not understand why the word static is mentioned here? Isn't the implicit object parameter the this pointer ( which only exists in non-static functions ) ?
Edit
in this link : link
quote :
The keyword this is a rvalue (until C++11)prvalue (since C++11) expression whose value is the address of the implicit object parameter (object on which the non-static member function is being called). It can appear in the following contexts:
It's useful to consider examples. When you have:
struct C {
void f(int);
void f(int) const;
};
C c;
c.f(42);
How does overload resolution pick? You effectively have a choice of:
// implicit object | regular
// parameter | parameter
void f(C&, int );
void f(C const&, int );
With the arguments (C, int). That ends up picking the first one, for being a better match.
Now, let's think of this example:
struct D {
static void g(int);
void g(long);
};
D d;
d.g(42);
Now, if we try to do the same thing:
// implicit object | regular
// parameter | parameter
void g(????????, int );
void g(D&, long );
We have two arguments, a D and an int. We don't know if we're going to call a static function or not yet, we still have to do overload resolution. How do we pick in this case? The non-static member function has an implicit object parameter, D&, but what do we do for the static one?
The C++ answer is we contrive a fake parameter, that is a perfect match for everything:
// implicit object | regular
// parameter | parameter
void g(contrived-match, int );
void g(D&, long );
And now, when we do overload resolution with (D, int), you can see that the static function is the best match (better conversion sequence for the second parameter).
Once we pick the static member function, we then ignore the object argument entirely. d.f(42) basically evaluates as D::f(42). But we didn't know that until we performed overload resolution - the contrived parameter exists to solve the problem of how to actually compare these cases.
This still applies even if there were just the one static member function - since d.f(42) does have two parameters: the d and the 42, so the language needs to handle the d somehow (the alternative could've been to simply disallow this syntax, requiring D::f(42) if you wanted to call a static member function, but that seems a lot less nice).
Consider what happens if you don't have this rule and have a static method and non-static method with the same (explicit) parameters. Then to the non-static method an additional implicit parameter (this) will be added, but not to the static method. This will make the list of parameters of both methods different and will allow to overload the static method with non-static method with the same explicit parameters.
First things first, there is a difference between implicit object parameter and this pointer. The former is a reference type while the latter is a keyword and is an rvalue of pointer type. For example for a const qualified non-static member function the implicit object parameter is of type const X& while the this pointer is of type const X*. While for a non-const nonstatic member function the implicit object parameter is of type X& and the this is of type X*. This can be confirmed here.
isn't Implicit object parameter the ( this ) pointer ( which ( the ( this ) pointer ) only works with non-static functions )
No, both static as well as non static member functions have an implicit object parameter for the purposes of overload resolution as can be seen from over.match.funcs#2 which states:
The set of candidate functions can contain both member and non-member functions to be resolved against the same argument list. So that argument and parameter lists are comparable within this heterogeneous set, a member function is considered to have an extra parameter, called the implicit object parameter, which represents the object for which the member function has been called. For the purposes of overload resolution, both static and non-static member functions have an implicit object parameter, but constructors do not.
(emphasis mine)
I am learning about classes in C++. I came across the following statement from the standard:
During overload resolution, non-static cv-qualified member function of class X is treated as a function that takes an implicit parameter of type lvalue reference to cv-qualified X if it has no ref-qualifiers or if it has the lvalue ref-qualifier. Otherwise (if it has rvalue ref-qualifier), it is treated as a function taking an implicit parameter of type rvalue reference to cv-qualified X.
The above statement seems to imply that for a const qualified non-static member function of a class X will have an implicit parameter of type const X&.
But then i also came across:
The type of this in a member function of class X is X* (pointer to X). If the member function is cv-qualified, the type of this is cv X* (pointer to identically cv-qualified X). Since constructors and destructors cannot be cv-qualified, the type of this in them is always X*, even when constructing or destroying a const object.
So according to the above second quote the implicit this parameter for a const qualified non-static member function of class Xhas type const X*.
My question is that why is there such a difference. I mean during overload resolution for a const qualfied nonstatic member function, why is the implicit parameter considered as a const X& and not simply a const X* which seems to be the actual type of this.
The implicit object parameter is not the same as this. this is a pointer referring to the object on which the member function was called while the implicit object parameter is the imagined first parameter of the member function, which is passed the object expression in the member function call (whats left of . in the member access expression) and so should be a reference parameter.
It wouldn't make sense to use a pointer for the implicit object parameter. It would make it impossible to overload the function on value category of the object expression. If a member function is &&-qualified, the implicit object parameter is a rvalue reference, so that if the object expression is a rvalue overload resolution correctly overload with a &-qualified member function.
So the implicit object parameter is const T& in your example, but this has type const T*. There is no contradiction.
Came across a proposal called "rvalue reference for *this" in clang's C++11 status page.
I've read quite a bit about rvalue references and understood them, but I don't think I know about this. I also couldn't find much resources on the web using the terms.
There's a link to the proposal paper on the page: N2439 (Extending move semantics to *this), but I'm also not getting much examples from there.
What is this feature about?
First, "ref-qualifiers for *this" is a just a "marketing statement". The type of *this never changes, see the bottom of this post. It's way easier to understand it with this wording though.
Next, the following code chooses the function to be called based on the ref-qualifier of the "implicit object parameter" of the function†:
// t.cpp
#include <iostream>
struct test{
void f() &{ std::cout << "lvalue object\n"; }
void f() &&{ std::cout << "rvalue object\n"; }
};
int main(){
test t;
t.f(); // lvalue
test().f(); // rvalue
}
Output:
$ clang++ -std=c++0x -stdlib=libc++ -Wall -pedantic t.cpp
$ ./a.out
lvalue object
rvalue object
The whole thing is done to allow you to take advantage of the fact when the object the function is called on is an rvalue (unnamed temporary, for example). Take the following code as a further example:
struct test2{
std::unique_ptr<int[]> heavy_resource;
test2()
: heavy_resource(new int[500]) {}
operator std::unique_ptr<int[]>() const&{
// lvalue object, deep copy
std::unique_ptr<int[]> p(new int[500]);
for(int i=0; i < 500; ++i)
p[i] = heavy_resource[i];
return p;
}
operator std::unique_ptr<int[]>() &&{
// rvalue object
// we are garbage anyways, just move resource
return std::move(heavy_resource);
}
};
This may be a bit contrived, but you should get the idea.
Note that you can combine the cv-qualifiers (const and volatile) and ref-qualifiers (& and &&).
Note: Many standard quotes and overload resolution explanation after here!
† To understand how this works, and why #Nicol Bolas' answer is at least partly wrong, we have to dig in the C++ standard for a bit (the part explaining why #Nicol's answer is wrong is at the bottom, if you're only interested in that).
Which function is going to be called is determined by a process called overload resolution. This process is fairly complicated, so we'll only touch the bit that is important to us.
First, it's important to see how overload resolution for member functions works:
§13.3.1 [over.match.funcs]
p2 The set of candidate functions can contain both member and non-member functions to be resolved against the same argument list. So that argument and parameter lists are comparable within this heterogeneous set, a member function is considered to have an extra parameter, called the implicit object parameter, which represents the object for which the member function has been called. [...]
p3 Similarly, when appropriate, the context can construct an argument list that contains an implied object argument to denote the object to be operated on.
Why do we even need to compare member and non-member functions? Operator overloading, that's why. Consider this:
struct foo{
foo& operator<<(void*); // implementation unimportant
};
foo& operator<<(foo&, char const*); // implementation unimportant
You'd certainly want the following to call the free function, don't you?
char const* s = "free foo!\n";
foo f;
f << s;
That's why member and non-member functions are included in the so-called overload-set. To make the resolution less complicated, the bold part of the standard quote exists. Additionally, this is the important bit for us (same clause):
p4 For non-static member functions, the type of the implicit object parameter is
“lvalue reference to cv X” for functions declared without a ref-qualifier or with the & ref-qualifier
“rvalue reference to cv X” for functions declared with the && ref-qualifier
where X is the class of which the function is a member and cv is the cv-qualification on the member function declaration. [...]
p5 During overload resolution [...] [t]he implicit object parameter [...] retains its identity since conversions on the corresponding argument shall obey these additional rules:
no temporary object can be introduced to hold the argument for the implicit object parameter; and
no user-defined conversions can be applied to achieve a type match with it
[...]
(The last bit just means that you can't cheat overload resolution based on implicit conversions of the object a member function (or operator) is called on.)
Let's take the first example at the top of this post. After the aforementioned transformation, the overload-set looks something like this:
void f1(test&); // will only match lvalues, linked to 'void test::f() &'
void f2(test&&); // will only match rvalues, linked to 'void test::f() &&'
Then the argument list, containing an implied object argument, is matched against the parameter-list of every function contained in the overload-set. In our case, the argument list will only contain that object argument. Let's see how that looks like:
// first call to 'f' in 'main'
test t;
f1(t); // 't' (lvalue) can match 'test&' (lvalue reference)
// kept in overload-set
f2(t); // 't' not an rvalue, can't match 'test&&' (rvalue reference)
// taken out of overload-set
If, after all overloads in the set are tested, only one remains, the overload resolution succeeded and the function linked to that transformed overload is called. The same goes for the second call to 'f':
// second call to 'f' in 'main'
f1(test()); // 'test()' not an lvalue, can't match 'test&' (lvalue reference)
// taken out of overload-set
f2(test()); // 'test()' (rvalue) can match 'test&&' (rvalue reference)
// kept in overload-set
Note however that, had we not provided any ref-qualifier (and as such not overloaded the function), that f1 would match an rvalue (still §13.3.1):
p5 [...] For non-static member functions declared without a ref-qualifier, an additional rule applies:
even if the implicit object parameter is not const-qualified, an rvalue can be bound to the parameter as long as in all other respects the argument can be converted to the type of the implicit object parameter.
struct test{
void f() { std::cout << "lvalue or rvalue object\n"; }
};
int main(){
test t;
t.f(); // OK
test().f(); // OK too
}
Now, onto why #Nicol's answer is atleast partly wrong. He says:
Note that this declaration changes the type of *this.
That is wrong, *this is always an lvalue:
§5.3.1 [expr.unary.op] p1
The unary * operator performs indirection: the expression to which it is applied shall be a pointer to an object type, or a pointer to a function type and the result is an lvalue referring to the object or function to which the expression points.
§9.3.2 [class.this] p1
In the body of a non-static (9.3) member function, 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*. [...]
There is an additional use case for the lvalue ref-qualifier form. C++98 has language that allows non-const member functions to be called for class instances that are rvalues. This leads to all kinds of weirdness that is against the very concept of rvalueness and deviates from how built-in types work:
struct S {
S& operator ++();
S* operator &();
};
S() = S(); // rvalue as a left-hand-side of assignment!
S& foo = ++S(); // oops, dangling reference
&S(); // taking address of rvalue...
Lvalue ref-qualifiers solve these problems:
struct S {
S& operator ++() &;
S* operator &() &;
const S& operator =(const S&) &;
};
Now the operators work like those of the builtin types, accepting only lvalues.
Let's say you have two functions on a class, both with the same name and signature. But one of them is declared const:
void SomeFunc() const;
void SomeFunc();
If a class instance is not const, overload resolution will preferentially select the non-const version. If the instance is const, the user can only call the const version. And the this pointer is a const pointer, so the instance cannot be changed.
What "r-value reference for this` does is allow you to add another alternative:
void RValueFunc() &&;
This allows you to have a function that can only be called if the user calls it through a proper r-value. So if this is in the type Object:
Object foo;
foo.RValueFunc(); //error: no `RValueFunc` version exists that takes `this` as l-value.
Object().RValueFunc(); //calls the non-const, && version.
This way, you can specialize behavior based on whether the object is being accessed via an r-value or not.
Note that you are not allowed to overload between the r-value reference versions and the non-reference versions. That is, if you have a member function name, all of its versions either use the l/r-value qualifiers on this, or none of them do. You can't do this:
void SomeFunc();
void SomeFunc() &&;
You must do this:
void SomeFunc() &;
void SomeFunc() &&;
Note that this declaration changes the type of *this. This means that the && versions all access members as r-value references. So it becomes possible to easily move from within the object. The example given in the first version of the proposal is (note: the following may not be correct with the final version of C++11; it's straight from the initial "r-value from this" proposal):
class X {
std::vector<char> data_;
public:
// ...
std::vector<char> const & data() const & { return data_; }
std::vector<char> && data() && { return data_; }
};
X f();
// ...
X x;
std::vector<char> a = x.data(); // copy
std::vector<char> b = f().data(); // move
At the moment I have a class defined similar to this:
class dummy{
public:
dummy(void(&func)(int))
: member{func}{}
void(&member)(int);
};
but I want to have member defined as a const function reference. I'm not sure exactly how to write this or if it is even possible.
P.S. PLEASE don't recommend me std::function I'm not oblivious to it's existence and have no objection to it, I just want to know whether something like this is doable.
The syntax:
Syntactically, you can achieve this through a type alias (or typedef):
using function_t = void(int); // or typedef void (function_t)(int);
class dummy {
public:
const function_t& member;
dummy(const function_t& func)
: member{func}{}
};
The semantics:
However, the "correct syntax" doesn't buy you anything: dummy is the same as
class dummy {
public:
function_t& member;
dummy(function_t& func)
: member{func}{}
};
(which, in turn is the same as the OP's definition) because the const qualifiers are ignored as per C++11 8.3.5/6:
The effect of a cv-qualifier-seq in a function declarator is not the same as adding cv-qualification on top of the function type. In the latter case, the cv-qualifiers are ignored. [ Note: a function type that has a cv-qualifier-seq is not a cv-qualified type; there are no cv-qualified function types. —end note ]
What about rvalues?
As far as I undestand (from a comment to user657267's answer) the motivation for taking the argument as reference to const was to enable passing rvalues (temporaries):
void f(int) { }
function_t make_function() { return f; }
dummy d1(f);
dummy d2(make_function());
However, this doesn't compile, not because of dummy, but because make_function returns a function which is forbiden by C++11 8.3.5/8
If the type of a parameter includes a type of the form “pointer to array of unknown bound of T” or “reference to array of unknown bound of T,” the program is ill-formed.99 Functions shall not have a return type of type array or function, although they may have a return type of type pointer or reference to such things. There shall be no arrays of functions, although there can be arrays of pointers to functions.
A natural "solution" would be returning a reference:
function& make_function() { return f; }
or
function&& make_function() { return f; }
In both cases, the type of the expression make_function() is an lvalue (which defies the purpose of dummy using reference to const to enable passing rvalues) as per C++11 5.2.2/10
A function call is an lvalue if the result type is an lvalue reference type or an rvalue reference to function type, an xvalue if the result type is an rvalue reference to object type, and a prvalue otherwise.
Actually, value category is not an issue here. With any legal declaration of make_function and any definition of dummy seem above the declarations of d1 and d2 work fine.
Nevertheless, the mentioned comment to user657267's answer talks about passing a lambda but a reference to function cannot bind to lambda expression because it has a different type:
dummy d3([](int){}); // Error!
The suggested solution
Instead of references use pointers:
using function_t = void(int); // or typedef void (function_t)(int);
class dummy {
public:
function_t* member;
dummy(function_t* func)
: member{func}{}
};
void f(int) { }
function_t& make_function() { return f; }
dummy d1(f); // OK
dummy d2(make_function()); // OK
dummy d3([](int){}); // OK
Final remarks:
Again, declaring const function_t* member and dummy(const function_t* func) doesn't buy you anything because, as per references, the const qualifiers are ignored.
The initialization of d1 and d2 work because functions are implicitly converted to pointer to functions (see C++ 4.3/1).
If a function argument is of function type, the compiler changes its type to a pointer to function. Hence, dummy's constructor can be declared as dummy(function_t func);.
The initialization of d3 works because captureless lambdas are implicitly converted to pointer to functions. It wouldn't work for lambdas with captures.
Came across a proposal called "rvalue reference for *this" in clang's C++11 status page.
I've read quite a bit about rvalue references and understood them, but I don't think I know about this. I also couldn't find much resources on the web using the terms.
There's a link to the proposal paper on the page: N2439 (Extending move semantics to *this), but I'm also not getting much examples from there.
What is this feature about?
First, "ref-qualifiers for *this" is a just a "marketing statement". The type of *this never changes, see the bottom of this post. It's way easier to understand it with this wording though.
Next, the following code chooses the function to be called based on the ref-qualifier of the "implicit object parameter" of the function†:
// t.cpp
#include <iostream>
struct test{
void f() &{ std::cout << "lvalue object\n"; }
void f() &&{ std::cout << "rvalue object\n"; }
};
int main(){
test t;
t.f(); // lvalue
test().f(); // rvalue
}
Output:
$ clang++ -std=c++0x -stdlib=libc++ -Wall -pedantic t.cpp
$ ./a.out
lvalue object
rvalue object
The whole thing is done to allow you to take advantage of the fact when the object the function is called on is an rvalue (unnamed temporary, for example). Take the following code as a further example:
struct test2{
std::unique_ptr<int[]> heavy_resource;
test2()
: heavy_resource(new int[500]) {}
operator std::unique_ptr<int[]>() const&{
// lvalue object, deep copy
std::unique_ptr<int[]> p(new int[500]);
for(int i=0; i < 500; ++i)
p[i] = heavy_resource[i];
return p;
}
operator std::unique_ptr<int[]>() &&{
// rvalue object
// we are garbage anyways, just move resource
return std::move(heavy_resource);
}
};
This may be a bit contrived, but you should get the idea.
Note that you can combine the cv-qualifiers (const and volatile) and ref-qualifiers (& and &&).
Note: Many standard quotes and overload resolution explanation after here!
† To understand how this works, and why #Nicol Bolas' answer is at least partly wrong, we have to dig in the C++ standard for a bit (the part explaining why #Nicol's answer is wrong is at the bottom, if you're only interested in that).
Which function is going to be called is determined by a process called overload resolution. This process is fairly complicated, so we'll only touch the bit that is important to us.
First, it's important to see how overload resolution for member functions works:
§13.3.1 [over.match.funcs]
p2 The set of candidate functions can contain both member and non-member functions to be resolved against the same argument list. So that argument and parameter lists are comparable within this heterogeneous set, a member function is considered to have an extra parameter, called the implicit object parameter, which represents the object for which the member function has been called. [...]
p3 Similarly, when appropriate, the context can construct an argument list that contains an implied object argument to denote the object to be operated on.
Why do we even need to compare member and non-member functions? Operator overloading, that's why. Consider this:
struct foo{
foo& operator<<(void*); // implementation unimportant
};
foo& operator<<(foo&, char const*); // implementation unimportant
You'd certainly want the following to call the free function, don't you?
char const* s = "free foo!\n";
foo f;
f << s;
That's why member and non-member functions are included in the so-called overload-set. To make the resolution less complicated, the bold part of the standard quote exists. Additionally, this is the important bit for us (same clause):
p4 For non-static member functions, the type of the implicit object parameter is
“lvalue reference to cv X” for functions declared without a ref-qualifier or with the & ref-qualifier
“rvalue reference to cv X” for functions declared with the && ref-qualifier
where X is the class of which the function is a member and cv is the cv-qualification on the member function declaration. [...]
p5 During overload resolution [...] [t]he implicit object parameter [...] retains its identity since conversions on the corresponding argument shall obey these additional rules:
no temporary object can be introduced to hold the argument for the implicit object parameter; and
no user-defined conversions can be applied to achieve a type match with it
[...]
(The last bit just means that you can't cheat overload resolution based on implicit conversions of the object a member function (or operator) is called on.)
Let's take the first example at the top of this post. After the aforementioned transformation, the overload-set looks something like this:
void f1(test&); // will only match lvalues, linked to 'void test::f() &'
void f2(test&&); // will only match rvalues, linked to 'void test::f() &&'
Then the argument list, containing an implied object argument, is matched against the parameter-list of every function contained in the overload-set. In our case, the argument list will only contain that object argument. Let's see how that looks like:
// first call to 'f' in 'main'
test t;
f1(t); // 't' (lvalue) can match 'test&' (lvalue reference)
// kept in overload-set
f2(t); // 't' not an rvalue, can't match 'test&&' (rvalue reference)
// taken out of overload-set
If, after all overloads in the set are tested, only one remains, the overload resolution succeeded and the function linked to that transformed overload is called. The same goes for the second call to 'f':
// second call to 'f' in 'main'
f1(test()); // 'test()' not an lvalue, can't match 'test&' (lvalue reference)
// taken out of overload-set
f2(test()); // 'test()' (rvalue) can match 'test&&' (rvalue reference)
// kept in overload-set
Note however that, had we not provided any ref-qualifier (and as such not overloaded the function), that f1 would match an rvalue (still §13.3.1):
p5 [...] For non-static member functions declared without a ref-qualifier, an additional rule applies:
even if the implicit object parameter is not const-qualified, an rvalue can be bound to the parameter as long as in all other respects the argument can be converted to the type of the implicit object parameter.
struct test{
void f() { std::cout << "lvalue or rvalue object\n"; }
};
int main(){
test t;
t.f(); // OK
test().f(); // OK too
}
Now, onto why #Nicol's answer is atleast partly wrong. He says:
Note that this declaration changes the type of *this.
That is wrong, *this is always an lvalue:
§5.3.1 [expr.unary.op] p1
The unary * operator performs indirection: the expression to which it is applied shall be a pointer to an object type, or a pointer to a function type and the result is an lvalue referring to the object or function to which the expression points.
§9.3.2 [class.this] p1
In the body of a non-static (9.3) member function, 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*. [...]
There is an additional use case for the lvalue ref-qualifier form. C++98 has language that allows non-const member functions to be called for class instances that are rvalues. This leads to all kinds of weirdness that is against the very concept of rvalueness and deviates from how built-in types work:
struct S {
S& operator ++();
S* operator &();
};
S() = S(); // rvalue as a left-hand-side of assignment!
S& foo = ++S(); // oops, dangling reference
&S(); // taking address of rvalue...
Lvalue ref-qualifiers solve these problems:
struct S {
S& operator ++() &;
S* operator &() &;
const S& operator =(const S&) &;
};
Now the operators work like those of the builtin types, accepting only lvalues.
Let's say you have two functions on a class, both with the same name and signature. But one of them is declared const:
void SomeFunc() const;
void SomeFunc();
If a class instance is not const, overload resolution will preferentially select the non-const version. If the instance is const, the user can only call the const version. And the this pointer is a const pointer, so the instance cannot be changed.
What "r-value reference for this` does is allow you to add another alternative:
void RValueFunc() &&;
This allows you to have a function that can only be called if the user calls it through a proper r-value. So if this is in the type Object:
Object foo;
foo.RValueFunc(); //error: no `RValueFunc` version exists that takes `this` as l-value.
Object().RValueFunc(); //calls the non-const, && version.
This way, you can specialize behavior based on whether the object is being accessed via an r-value or not.
Note that you are not allowed to overload between the r-value reference versions and the non-reference versions. That is, if you have a member function name, all of its versions either use the l/r-value qualifiers on this, or none of them do. You can't do this:
void SomeFunc();
void SomeFunc() &&;
You must do this:
void SomeFunc() &;
void SomeFunc() &&;
Note that this declaration changes the type of *this. This means that the && versions all access members as r-value references. So it becomes possible to easily move from within the object. The example given in the first version of the proposal is (note: the following may not be correct with the final version of C++11; it's straight from the initial "r-value from this" proposal):
class X {
std::vector<char> data_;
public:
// ...
std::vector<char> const & data() const & { return data_; }
std::vector<char> && data() && { return data_; }
};
X f();
// ...
X x;
std::vector<char> a = x.data(); // copy
std::vector<char> b = f().data(); // move