Why can't a const method return a non-const reference?

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Why won't the method getRanks() below compile, and how can I fix it gracefully?

All I want do is define a member accessor method that returns a reference to a member. The reference is not const since I might well modify what it refers to later. But since the member method does not modify the object, I declare it const. The compiler (clang, std=c++11) then insists that there is a "binding of reference" that "drops qualifiers". But I'm NOT dropping qualifiers, am I? And if I am, why:

struct teststruct{
  vector<int> ranks;
  vector<int>& getRanks()const{
    return ranks;

Now, the code compiles if I change the return statement to cast away the const:

return const_cast<vector<int>&>(ranks);

But "ranks" should not be const in the first place, I don't see why I need to const_cast the const away. I don't even know if it's safe to do this.

Anyway, is there a cleaner to write this method? Can someone explain why such a simple common-sense method fails? I do want to declare the getRanks() method "const" so that I can call it from other const methods.

The idea behind the const member function is you should be able to call them on const objects. const functions can't modify the object.

Say you have a class

class A
   int data;
   void foo() const

and on object and a function call:

A const a;

Inside A::foo, this->data is treated as if its type is int const, not int. Hence, you are not able to modify this->data in A:foo().

Coming to your example, the type of this->ranks in getRanks() is to be considered as const vector<int> and not vector<int>. Since, auto conversion of const vector<int> to vector<int>& is not allowed, the compiler complains when you define the function as:

vector<int>& getRanks()const{
    return ranks;

It won't complain if you define the function as:

const vector<int>& getRanks()const{
    return ranks;

since const vector<int> can be auto converted to const vector<int>&.

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ranks is const because the enclosing object (*this) is const, so you have to return a reference to a std::vector<int> const.

If you want to allow the client to modify the vector (and thereby affecting the member), then the getter should not be const. Note that the getter is silly anyway, since ranks is already a public data member.

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You are returning a reference to ranks, which is a member of teststruct. That means that anybody that gets this reference could modify the internals of the teststruct object. So the const is a lie.

Don't cast away the const. Instead, decide between whether you want the function to be const and return a copy of ranks or const reference, or to be non-const and return a mutable reference. You can always have both if necessary.

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It drops qualifiers because you return a reference to ranks. Any code after can modify ranks. Ex:

auto v = teststruct.getRanks();
v[1] = 5; //assuming v.size() > 1

You can fix this by returning a copy:

vector<int> getRanks() const

Or a const reference:

const vector<int>& getRanks() const

If you want ranks changable even in a const object, you could do this:

mutable vector<int> ranks;

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  • teststruct const x; x.getRanks().emplace_back(); -> UB.
  • If getter is not const, it cannot be called from a const method. But I want to do do that, I want to call it sometimes from const method, sometimes from non_const methods.
  • @kdog then make two getters, one const and one non-const.
  • A la std::vector::operator[].
  • But hold on. I thought a "const" just before the semicolon in the declaration of a method just meant that that method ITSELF did not modify the object of which it is a member function. That's what the Stroustrup book says if I recall. I am not declaring that any usage from now into the future of the returned will not modify the object. I am just declaring that the function itself will not. And I DO wan t to return a reference that can be modified, not a copy! Also is the const_cast idea safe to use?
  • @kdog Well this would allow the method to indirectly cause the object to be modified. The const_cast idea is not safe, because the compiler can make assumptions based on the fact it believes that it won't be modified.
  • Rather than "the const is a lie", I'd say "the const/return type breaks const correctness".
  • No, the compiler should give an error at the v.push_back(42) line, not the getRanks() line...although I am beginning to see the issue. Why is this so confusing and counterintuitive to me and obvious to everyone else ;-)
  • @kdog The const for the member function means that the object that the function is called on is const. In other words the this-pointer is const. And because this is const, this->anything is also const. There is no requirement for getRanks to return a const &, but there is a requirement to return an object with the return type's type.
  • No this doesn't fix it because I I might want later to modify the reference returned by getRanks(). Do I need two getRanks, one const and one not? I just want to declare that getRanks() ITSELF does not modify object. That it can be called in a const method.
  • No I do not want ranks to be mutable , I do want const checking.
  • Then it sounds like two overloads vector<int>& getRanks(); and const vector<int>& getRanks() const; is what you want.