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Why can't a destructor have reference qualifiers?

By : Silas
Date : August 01 2020, 10:00 AM
I hope this helps you . First, there must be only one destructor per class. Allowing ref-qualifiers on the destructor would make it possible to overload the destructor.
Another possible reason is to be consistent with const and volatile qualifiers:
code :

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Template Reference Collapsing Dropping cv-qualifiers for const Reference Return Type

By : user3513771
Date : March 29 2020, 07:55 AM
Hope that helps You could use std::remove_reference to drop the reference and add it back:
code :
#include <type_traits>

template<typename T>
class Raw
    T obj;
    using NR = std::remove_reference_t<T>;

    Raw() {};
    Raw(const T& init): obj(init) {};

    NR& get() {return obj;};
    const NR& get() const {return obj;};

What are reference qualifiers for functions?

By : user3865130
Date : March 29 2020, 07:55 AM
hop of those help? According to this webpage, a non-static member function can have a trailing & or && in its declaration. They have the following example
code :
struct S {
  virtual int f(char) const, g(int) &&;

struct D : S {
  virtual int f(char) const override;
  virtual int g(int) && override;

Binding to reference of type discards qualifiers

By : Coolieo
Date : March 29 2020, 07:55 AM
seems to work fine std::map makes its keys immutable, so references to mutable versions of its keys is not possible without violating const-correctness. If you are certain that this version of a map where you're doing lookups by value is correct, then you'll need to roll your own data structure to make this code work.
Alternatively, because I have difficulty believing that that's actually a correct version of your program, consider the following: in all instances where you look-up values in edges, you're searching for a std::string associated with two Tnode objects. It seems pretty clear that the correct data structure for this code is not std::map>, but std::map, Tedge>. Tnode in this case represents half of the key looking up a Tedge object, which is how you use it.
code :
void add(Tnode n, Tnode n2, Tedge e) {
    edges.insert(std::make_pair(std::make_pair(n, n2), e));

bool has(Tedge edge) const{
    bool l = false;
    for(typename std::map<Tedge, std::pair<Tnode, Tnode>>::const_iterator it = edges.begin(); it != edges.end() && !l; it++) {
        l = it -> second == edge;

    return l;

const Tedge operator()(const Tnode &first, const Tnode &sec) const {
    std::map<std::pair<Tnode, Tnode>, Tedge>::const_iterator it = edges.find(std::pair<Tnode, Tnode>(first, sec));
    if(it != edges.end()) 
        return it-> second;
        throw std::runtime_error("No such element found");

Tedge & operator()(const Tnode &first, const Tnode &sec) {
    std::map<std::pair<Tnode, Tnode>, Tedge>::iterator it = edges.find(std::pair<Tnode, Tnode>(first, sec));
    if(it != edges.end()) 
        return it-> second;
        throw std::runtime_error("No such element found");
void add(Tnode n, Tnode n2, Tedge e) {
    edges.emplace(std::make_pair(n, n2), e);

bool has(Tedge const& edge) const{
    for(auto const& entry : edges) 
        if(entry.second == edge) 
            return true;
    return false;

const Tedge operator()(const Tnode &first, const Tnode &sec) const {
    if(auto it = edges.find(std::make_pair(first, sec)); it != edges.end())
        return it->second;
        throw std::runtime_error("No such element found");

Tedge & operator()(const Tnode &first, const Tnode &sec) {
    if(auto it = edges.find(std::make_pair(first, sec)); it != edges.end())
        return it->second;
        throw std::runtime_error("No such element found");

Why derived class destructor called even though base class destructor is not virtual if object is created as reference

By : Mohamed Hassan
Date : March 29 2020, 07:55 AM
Any of those help This is a special case when const references to temporaries are involved. The destructor for the temporary is correctly invoked, rather than the destructor of the reference since, after all, the lifetime of the temporary is merely extended.

C++98/03 reference collapsing and cv qualifiers

By : Bruno
Date : March 29 2020, 07:55 AM
To fix this issue GCC 4.7.2 does not compile this when the flag -std=c++98 is specified. In fact, in C++98 (as well as in C++03) references to references do not collapse.
An attempt to instantiate f, where T = int&, produces the following function signature (here I intentionally switch the position of the argument type T and the const specifier, which is allowed because const T& is the same as T const&):
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