如何在不添加/删除 mem 函数的“const”性的情况下重载 getter 函数？

Question

我想不通，为什么 C++ 不允许根据 return 类型进行重载，如下例所示三个成员（getter）函数具有不同的函数签名甚至何时存储指向成员函数的指针，我们需要不同的内存函数指针类型，例如：

for instance T = std::string
using constRefPtr   = const std::string&(MyStruct::*)() const;
using constValuePtr = const std::string(MyStruct::*)() const;
using valuePtr      = std::string(MyStruct::*)() const;

我读过这个 similar post，其中建议使用 const 和 non-cost 成员函数。

问题：如何在不删除 constness 的情况下使以下 (getter) 重载工作每个成员函数（如果可以通过标准 C++）？

我正在使用 C++17。

#include <iostream>
#include <string>

template<typename T> class MyStruct
{
    T m_val;
public: 
    explicit MyStruct(const T& value) 
        : m_val(value)
    {}
    const T& getVal() const {   return m_val; } // get val as const ref(no copy of member)
    const T getVal() const  {   return m_val; } // get a const member as return
    T getVal() const        {   return m_val; } // get a copy of member
};

int main()
{
    MyStruct<std::string> obj{"string"};
    const auto& val_const_ref = obj.getVal();  // overload const std::string& getVal() const
    const auto val_const = obj.getVal();       // overload const std::string getVal()  const
    auto val = obj.getVal();                   // overload std::string getVal()  const
    return 0;
}

我收到的错误信息：

error C2373 : 'MyStruct<T>::getVal' : redefinition; different type modifiers
note: see declaration of 'MyStruct<T>::getVal'
note: see reference to class template instantiation 'MyStruct<T>' being compiled
error C2059 : syntax error : 'return'
error C2238 : unexpected token(s) preceding ';'
error C2143 : syntax error : missing ';' before '}'
error C2556 : 'const T MyStruct<T>::getVal(void) const' : overloaded function differs only by return type from 'const T &MyStruct<T>::getVal(void) const'
1 > with
1 > [
    1 > T = std::string
        1 > ]
    1 > C:\Z Drive\CPP Programs\Visual Studio Project\Main.cc(62) : note: see declaration of 'MyStruct<std::string>::getVal'
note: see reference to class template instantiation 'MyStruct<std::string>' being compiled
error C2373 : 'MyStruct<std::string>::getVal' : redefinition; different type modifiers
note: see declaration of 'MyStruct<std::string>::getVal'
error C2059 : syntax error : 'return'
error C2238 : unexpected token(s) preceding ';'
error C2146 : syntax error : missing ';' before identifier 'T'
error C2530 : 'val_const_ref' : references must be initialized
error C2789 : 'val_const' : an object of const - qualified type must be initialized
note: see declaration of 'val_const'

Answer 1

Question: How could I make the following (getter)overloads work without removing constness of each member functions(if it is possible through standard C++)?

适当地命名它们。正如@Slava 指出的那样，当 return 复制值时，区分 T 和 const T 毫无意义，所以类似：

const T& getConstRefVal() const { return m_val; } // get val as const ref(no copy of member)
T        getVal()         const { return m_val; } // get a copy of member

Answer 2

你只是...不能在 return 类型上超载，句号。

您可以创建两个不同名称的函数：

T const& ref() const { return m_val; }
T val() const        { return m_val; }

哪些自身可以根据 constness 或 &ness 重载：

T const& ref() const { return m_val; }
T&       ref()       { return m_val; }

T val() const&       { return m_val; }
T val() &&           { return std::move(m_val); }

Answer 3

这是不可能的。您不能重载 return 类型。重载解析考虑了函数签名。函数签名由以下部分组成：

函数名称
cv-qualifiers
参数类型

标准说：

1.3.11签名

the information about a function that participates in overload resolution (13.3): its parameter-type-list (8.3.5) and, if the function is a class member, the cv-qualifiers (if any) on the function itself and the class in which the member function is declared. [...]

（根据 Luchian Grigore's answer 稍作编辑）

如何在不添加/删除 mem 函数的“const”性的情况下重载 getter 函数？

How to overload getter functions without adding/ removing `const`ness of mem function?

c++

overloading

member-functions

c++17