This question already has answers here:
What does the single ampersand after the parameter list of a member function declaration mean?
(3 answers)
What is "rvalue reference for *this"?
(3 answers)
Closed 6 years ago.
Looking at Boost::Optional optional class template header I come across this:
T const& operator*() const&
T& operator*() &;
T&& operator*() &&;
For the life of me I can't find this syntax anywhere else (a reference as the last symbol) I would assume it has something to do with overloading on the type(const l-val, l-val, r-val) of the object the operator belongs to, but I haven't seen this described anywhere.
Could someone tell me what this syntax means?
Related
This question already has answers here:
C++ Return value, reference, const reference
(5 answers)
What is a reference variable in C++?
(12 answers)
Closed 1 year ago.
I am trying to understand the useage of 'const' and '&' in the following function declaration. I know that the last 'const' means the function cannot change member variables in the class and that 'const std::string& message' means the variable passed to the function cannot be changed, but I don't understand the meaning of 'const Logger&'. What is the purpose of this first 'const' and why is there an '&' following 'Logger'? Is this function meant to return an address or a pointer?
const Logger& log(const std::string& message) const;
So const Logger& is the return type of log. const means you will not be able to edit the return value at all. The return type Logger& means you'll get a reference to a Logger and not a copy of it.
This question already has answers here:
ampersand (&) at the end of variable etc
(5 answers)
Closed 1 year ago.
What is the return type of such a function(this function is the member of a class)?
class_name& operator =(const class_name& rightSide)
The return type is class_name&
This question already has answers here:
Overload resolution between object, rvalue reference, const reference
(1 answer)
C++ function overloading resolution involving pass-by-value, reference and constant reference
(3 answers)
Function Overloading Based on Value vs. Const Reference
(6 answers)
Closed 2 years ago.
How could i select the overloaded function that i mean to call ?
Consider this code:
void foo (std::vector<int> const &variable);
void foo (std::vector<int> variable);
For example in above code i want to call void foo (std::vector<int>);, I tried :
void bar ()
{
std::vector<int> tmp;
foo(tmp);
foo(static_cast<std::vector<int>(tmp));
foo(boost::implicit_cast<std::vector<int>>(tmp));
foo((std::vector<int>)tmp);
foo(std::vector<int>(tmp));
}
But it's failed, How could i do that without changing the functions signature ?
This question already has answers here:
Why are rvalues references variables not rvalue?
(3 answers)
Rvalue Reference is Treated as an Lvalue?
(4 answers)
Closed 3 years ago.
Considering this code
class T {
public:
T(T& x) = delete;
T(T&& x) {}
};
void foo(T&& b) {
T y(b);
}
I was expecting that b; which is an rvalue by declaration; and seemingly usage, should be passed into the move constructor of T in foo().
Instead; I get a compilation error reporting that T& has been deleted.
Replacing it with
void foo(T&& c) {
T y(std::move(c));
}
Results in the expected success; but obviously one doesn't want to litter their code with std::move everywhere.
As tempting as it is to blame visual studio - in this case I suspect it's my understanding that's wrong. Can someone please explain why move constructor isn't used?
This question already has answers here:
Meaning of 'const' last in a function declaration of a class?
(12 answers)
Closed 8 years ago.
In the book "Programming Interviews Exposed" by John Mongan (see page 33), theres a function declaration for a Singly Linked List Element:
const T& value() const {...}
I understand what everything before the method name means, namely that it is a reference to a template datatype which you may not modify, but what does the extra const after the value() signify? A constant reference? I thought references were already constant (i.e. unchangeable alias which is the object).
const T& value() const {...}
This means the function value() will return a const T& type and in between (in the function) won't modify the class itself.
Say I write:
class Cfoo
{
void foo() const
{
//Cfoo will not be modified here
}
}
If I directly quote from MS Docs:
Declaring a member function with the const keyword specifies that the function is a "read-only" function that does not modify the object for which it is called. A constant member function cannot modify any non-static data members or call any member functions that aren't constant.
The const after the method name signifies that the method does not modify any field of the object. Therefore, it can be invoked on a const instance of the object.