我可以在覆盖虚函数的返回类型中丢失“constness”吗？

Question

The following code compiles and runs, and no warning is emitted by either gcc or clang:

#include <iostream>

struct Base {
    virtual ~Base() = default;
    virtual std::string const& get() = 0;
};

struct Derived: Base {
    virtual std::string& get() override { return m; }
    std::string m;
};

int main()
{
    Derived d;
    d.get() = "Hello, World";

    Base& b = d;
    std::cout << b.get() << "\n";
}

Is std::string& covariant with std::string const& then?

Answer 1

Yes

This is specified in class.virtual , in the latest draft (n4606) we see:

§10.3 7/ The return type of an overriding function shall be either identical to the return type of the overridden function or covariant with the classes of the functions. If a function D::f overrides a function B::f , the return types of the functions are covariant if they satisfy the following criteria:

• both are pointers to classes, both are lvalue references to classes, or both are rvalue references to classes ¹¹¹
• the class in the return type of B::f is the same class as the class in the return type of D::f , or is an unambiguous and accessible direct or indirect base class of the class in the return type of D::f
• both pointers or references have the same cv-qualification and the class type in the return type of D::f has the same cv-qualification as or less cv-qualification than the class type in the return type of B::f .

Specifically, the last point addresses exactly the case here: it is acceptable for an overriding type to lose the const and/or volatile qualifiers (it cannot, however, gain them).

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Note: as mentioned by @george above, paragraph 8/ used to prevent this from working with incomplete class types, but this was since fixed .