抽象类:成员函数virtual..的抽象返回类型无效

Question

在我的程序中，我有这样的类层次:

#include <iostream>
#include <cmath>
#include <sstream>
using namespace std;
class aa;
class bb;
class root
{
public:
    virtual ~root() {}
    virtual root add(const aa& a) const=0;
    virtual root add(const bb& a) const=0;
};
class aa: public root
{
public:
    aa() { }
    aa(const aa& a) { }
    virtual root add(const aa& a) const
    { return root(new aa()); }
    virtual root add(const bb& a) const
    { return root(new bb()); }
};
class bb: public root
{
public:
    bb() { }
    bb(const bb& b) {}
    virtual root add(const aa& a) const
    { return root(new bb()); }
    virtual root add(const bb& a) const
    { return root(new bb()); }
};
int main(int argc, char **argv)
{
}

，但我仍然得到错误的编译过程。我不能改变我的类层次结构，但有可能使我想要的东西在这里?

编辑类:

#include <iostream>
#include <cmath>
#include <sstream>
using namespace std;
class root
{
public:
    virtual ~root() {}
    virtual root add(const root& a) const=0;
    virtual root add(const root& b) const=0;
};
class aa: public root
{
public:
    aa() { }
    aa(const aa& a) { }
    virtual root add(const root& a) const
    { return root(new aa()); }
    virtual root add(const root& b) const
    { return root(new bb()); }
};
class bb: public root
{
public:
    bb() { }
    bb(const bb& b) {}
    virtual root add(const root& a) const
    { return root(new bb()); }
    virtual root add(const root& b) const
    { return root(new bb()); }
};
int main(int argc, char **argv)
{
}

编辑类错误:

/home/brian/Desktop/Temp/Untitled2.cpp|11|error: ‘virtual root root::add(const root&) const’ cannot be overloaded|
/home/brian/Desktop/Temp/Untitled2.cpp|10|error: with ‘virtual root root::add(const root&) const’|
/home/brian/Desktop/Temp/Untitled2.cpp|10|error: invalid abstract return type for member function ‘virtual root root::add(const root&) const’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   because the following virtual functions are pure within ‘root’:|
/home/brian/Desktop/Temp/Untitled2.cpp|10|note:     virtual root root::add(const root&) const|
/home/brian/Desktop/Temp/Untitled2.cpp|11|error: invalid abstract return type for member function ‘virtual root root::add(const root&) const’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   since type ‘root’ has pure virtual functions|
/home/brian/Desktop/Temp/Untitled2.cpp|20|error: invalid abstract return type for member function ‘virtual root aa::add(const root&) const’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   since type ‘root’ has pure virtual functions|
/home/brian/Desktop/Temp/Untitled2.cpp|22|error: invalid abstract return type for member function ‘virtual root aa::add(const root&) const’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   since type ‘root’ has pure virtual functions|
/home/brian/Desktop/Temp/Untitled2.cpp|22|error: ‘virtual root aa::add(const root&) const’ cannot be overloaded|
/home/brian/Desktop/Temp/Untitled2.cpp|20|error: with ‘virtual root aa::add(const root&) const’|
/home/brian/Desktop/Temp/Untitled2.cpp||In member function ‘virtual root aa::add(const root&) const’:|
/home/brian/Desktop/Temp/Untitled2.cpp|20|error: invalid abstract return type for member function ‘virtual root aa::add(const root&) const’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   since type ‘root’ has pure virtual functions|
/home/brian/Desktop/Temp/Untitled2.cpp|21|error: cannot allocate an object of abstract type ‘root’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   since type ‘root’ has pure virtual functions|
/home/brian/Desktop/Temp/Untitled2.cpp||In member function ‘virtual root aa::add(const root&) const’:|
/home/brian/Desktop/Temp/Untitled2.cpp|22|error: invalid abstract return type for member function ‘virtual root aa::add(const root&) const’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   since type ‘root’ has pure virtual functions|
/home/brian/Desktop/Temp/Untitled2.cpp|23|error: cannot allocate an object of abstract type ‘root’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   since type ‘root’ has pure virtual functions|
/home/brian/Desktop/Temp/Untitled2.cpp|23|error: expected primary-expression before ‘(’ token|
/home/brian/Desktop/Temp/Untitled2.cpp|23|error: expected type-specifier before ‘bb’|
/home/brian/Desktop/Temp/Untitled2.cpp|23|error: expected ‘)’ before ‘bb’|
/home/brian/Desktop/Temp/Untitled2.cpp|32|error: invalid abstract return type for member function ‘virtual root bb::add(const root&) const’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   since type ‘root’ has pure virtual functions|
/home/brian/Desktop/Temp/Untitled2.cpp|34|error: invalid abstract return type for member function ‘virtual root bb::add(const root&) const’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   since type ‘root’ has pure virtual functions|
/home/brian/Desktop/Temp/Untitled2.cpp|34|error: ‘virtual root bb::add(const root&) const’ cannot be overloaded|
/home/brian/Desktop/Temp/Untitled2.cpp|32|error: with ‘virtual root bb::add(const root&) const’|
/home/brian/Desktop/Temp/Untitled2.cpp||In member function ‘virtual root bb::add(const root&) const’:|
/home/brian/Desktop/Temp/Untitled2.cpp|32|error: invalid abstract return type for member function ‘virtual root bb::add(const root&) const’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   since type ‘root’ has pure virtual functions|
/home/brian/Desktop/Temp/Untitled2.cpp|33|error: cannot allocate an object of abstract type ‘root’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   since type ‘root’ has pure virtual functions|
/home/brian/Desktop/Temp/Untitled2.cpp||In member function ‘virtual root bb::add(const root&) const’:|
/home/brian/Desktop/Temp/Untitled2.cpp|34|error: invalid abstract return type for member function ‘virtual root bb::add(const root&) const’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   since type ‘root’ has pure virtual functions|
/home/brian/Desktop/Temp/Untitled2.cpp|35|error: cannot allocate an object of abstract type ‘root’|
/home/brian/Desktop/Temp/Untitled2.cpp|6|note:   since type ‘root’ has pure virtual functions|
||=== Build finished: 38 errors, 0 warnings ===|

Answer 1

不能按值返回root，因为root是抽象的，因此永远不可能存在任何类型的root的值。

你可能想返回一个指针:

#include <memory>
std::unique_ptr<root> do_you_feel_lucky(aa const & x, bb const & y)
{
    if (rand() % 2 == 0)
        return { new aa(x) };
    else
        return { new bb(y) };
}

你所拥有的感觉很像一个"克隆"或"虚拟复制"功能:

struct Base
{
    virtual std::unique_ptr<Base> clone() const = 0;
};
struct Derived : Base
{
    virtual std::unique_ptr<Base> clone() const
    {
        return { new Derived(*this); }
    }
};

既然你问了关于引用的问题，那么你还可以做另一件事，尽管它看起来有点没有意义:从几个派生对象中选择一个引用，并返回一个基引用。

root & pick_one_from_two(aa & x, bb & y)
{
    return rand() % 2 == 0 ? x : y;
}