c++ - C++ 中另一个带有闭包的抽象类问题

Question

我在使用g++ -std=c++11参数编译此代码时遇到问题：

#include <iostream>
#include <unordered_map>
#include <functional>
#include <iostream>
using namespace std;;

template <typename T> class opt {
    public:
        opt() {};
        ~opt() {};
        virtual int isEmpty() = 0;
        virtual T getObject() = 0;
};

template <typename T> class oopt : public opt<T> {
    private:
        T ret;
    public:
        oopt(T obj) { ret = obj; };
        ~oopt() override;
        int isEmpty() { return 0; };
        T getObject() { return ret; };
};
int main(void) {

    function<opt<int>(int)> fu = [](int j) { return (oopt<int>(10)); };

    return 1;
}

此代码返回以下错误：

assoc_array.cc: In instantiation of ‘class oopt<int>’:
assoc_array.cc:68:63:   required from here
assoc_array.cc:29:3: error: ‘oopt<T>::~oopt() [with T = int]’ marked override, but does not override
In file included from assoc_array.cc:3:0:
/usr/include/c++/4.7/functional: In instantiation of ‘static _Res std::_Function_handler<_Res(_ArgTypes ...), _Functor>::_M_invoke(const std::_Any_data&, _ArgTypes ...) [with _Res = opt<int>; _Functor = main()::<lambda(int)>; _ArgTypes = {int}]’:
/usr/include/c++/4.7/functional:2298:6:   required from ‘std::function<_Res(_ArgTypes ...)>::function(_Functor, typename std::enable_if<(! std::is_integral<_Functor>::value), std::function<_Res(_ArgTypes ...)>::_Useless>::type) [with _Functor = main()::<lambda(int)>; _Res = opt<int>; _ArgTypes = {int}; typename std::enable_if<(! std::is_integral<_Functor>::value), std::function<_Res(_ArgTypes ...)>::_Useless>::type = std::function<opt<int>(int)>::_Useless]’
assoc_array.cc:68:67:   required from here
/usr/include/c++/4.7/functional:1909:7: error: invalid abstract return type for function ‘static _Res std::_Function_handler<_Res(_ArgTypes ...), _Functor>::_M_invoke(const std::_Any_data&, _ArgTypes ...) [with _Res = opt<int>; _Functor = main()::<lambda(int)>; _ArgTypes = {int}]’
assoc_array.cc:7:29: note:   because the following virtual functions are pure within ‘opt<int>’:
assoc_array.cc:11:15: note:     int opt<T>::isEmpty() [with T = int]
assoc_array.cc:12:13: note:     T opt<T>::getObject() [with T = int]
In file included from assoc_array.cc:3:0:
/usr/include/c++/4.7/functional:1912:40: error: cannot allocate an object of abstract type ‘opt<int>’
assoc_array.cc:7:29: note:   since type ‘opt<int>’ has pure virtual functions
In file included from assoc_array.cc:3:0:
/usr/include/c++/4.7/functional:1743:2: error: ‘static _Functor* std::_Function_base::_Base_manager<_Functor>::_M_get_pointer(const std::_Any_data&) [with _Functor = main()::<lambda(int)>]’, declared using local type ‘main()::<lambda(int)>’, is used but never defined [-fpermissive]

使用带有静态转换到接口的闭包的错误在哪里？我已经阅读了类似的回复（抽象类中的 C++ 和继承，C++ - 派生类中的“未声明成员函数”），但如果我删除闭包定义，编译时似乎没有错误。提前致谢。

编辑

即使使用以下代码：

#include <iostream>
#include <unordered_map>
#include <functional>
#include <iostream>
using namespace std;;

template <typename T> class opt {
    public:
        opt<T>(T) {};
        ~opt<T>() {};
        virtual int isEmpty() = 0;
        virtual T getObject() = 0;
};

template <typename T> class oopt : public opt<T> {
    private:
        T ret;
    public:
        oopt<T>(T obj) { ret = obj; };
        ~oopt<T>() override;
        int isEmpty() { return 0; };
        T getObject() { return ret; };
};
int main(void) {

    function<opt<int>(int)> fu = [](int j) { return (oopt<int>(10)); };

    return 1;
}

我仍然有以下编译问题：

assoc_array.cc: In instantiation of ‘class oopt<int>’:
assoc_array.cc:26:64:   required from here
assoc_array.cc:20:4: error: ‘oopt<T>::~oopt() [with T = int]’ marked override, but does not override
assoc_array.cc: In instantiation of ‘oopt<T>::oopt(T) [with T = int]’:
assoc_array.cc:26:64:   required from here
assoc_array.cc:19:19: error: no matching function for call to ‘opt<int>::opt()’
assoc_array.cc:19:19: note: candidates are:
assoc_array.cc:9:4: note: opt<T>::opt(T) [with T = int]
assoc_array.cc:9:4: note:   candidate expects 1 argument, 0 provided
assoc_array.cc:7:30: note: constexpr opt<int>::opt(const opt<int>&)
assoc_array.cc:7:30: note:   candidate expects 1 argument, 0 provided
In file included from assoc_array.cc:3:0:
/usr/include/c++/4.7/functional: In instantiation of ‘static _Res std::_Function_handler<_Res(_ArgTypes ...), _Functor>::_M_invoke(const std::_Any_data&, _ArgTypes ...) [with _Res = opt<int>; _Functor = main()::<lambda(int)>; _ArgTypes = {int}]’:
/usr/include/c++/4.7/functional:2298:6:   required from ‘std::function<_Res(_ArgTypes ...)>::function(_Functor, typename std::enable_if<(! std::is_integral<_Functor>::value), std::function<_Res(_ArgTypes ...)>::_Useless>::type) [with _Functor = main()::<lambda(int)>; _Res = opt<int>; _ArgTypes = {int}; typename std::enable_if<(! std::is_integral<_Functor>::value), std::function<_Res(_ArgTypes ...)>::_Useless>::type = std::function<opt<int>(int)>::_Useless]’
assoc_array.cc:26:68:   required from here
/usr/include/c++/4.7/functional:1909:7: error: invalid abstract return type for function ‘static _Res std::_Function_handler<_Res(_ArgTypes ...), _Functor>::_M_invoke(const std::_Any_data&, _ArgTypes ...) [with _Res = opt<int>; _Functor = main()::<lambda(int)>; _ArgTypes = {int}]’
assoc_array.cc:7:30: note:   because the following virtual functions are pure within ‘opt<int>’:
assoc_array.cc:11:16: note:     int opt<T>::isEmpty() [with T = int]
assoc_array.cc:12:14: note:     T opt<T>::getObject() [with T = int]
In file included from assoc_array.cc:3:0:
/usr/include/c++/4.7/functional:1912:40: error: cannot allocate an object of abstract type ‘opt<int>’
assoc_array.cc:7:30: note:   since type ‘opt<int>’ has pure virtual functions
In file included from assoc_array.cc:3:0:
/usr/include/c++/4.7/functional:1743:2: error: ‘static _Functor* std::_Function_base::_Base_manager<_Functor>::_M_get_pointer(const std::_Any_data&) [with _Functor = main()::<lambda(int)>]’, declared using local type ‘main()::<lambda(int)>’, is used but never defined [-fpermissive]

score 1 · Accepted Answer

由于 gcc 是周围的标准编译器，他的错误不是最清楚的，甚至用 C++ 比 C 更严重，所以我通常用 clang 和 gcc 编译所有东西以确保。用 clang 编译：我用 aaronman 代码得到了这个，t

clang++ -std=c++11 test.cc                                                                                                                                                                              ~
test.cc:20:20: error: only virtual member functions can be marked 'override'
        ~oopt<T>() override;
                   ^~~~~~~~
test.cc:26:54: note: in instantiation of template class 'oopt<int>' requested here
    function<opt<int>(int)> fu = [](int j) { return (oopt<int>(10)); };
                                                     ^
test.cc:19:9: error: constructor for 'oopt<int>' must explicitly initialize the base class 'opt<int>' which does not have a default constructor
        oopt<T>(T obj) { ret = obj; };
        ^
test.cc:26:54: note: in instantiation of member function 'oopt<int>::oopt' requested here
    function<opt<int>(int)> fu = [](int j) { return (oopt<int>(10)); };
                                                     ^
test.cc:7:29: note: 'opt<int>' declared here
template <typename T> class opt {
                            ^
2 errors generated.

这是你的第一个：

clang++ -std=c++11 test2.cc                                                                                                                                                                             ~
test2.cc:20:17: error: only virtual member functions can be marked 'override'
        ~oopt() override;
                ^~~~~~~~
test2.cc:26:54: note: in instantiation of template class 'oopt<int>' requested here
    function<opt<int>(int)> fu = [](int j) { return (oopt<int>(10)); };
                                                     ^
1 error generated.

修复这些错误后，clang 在链接过程中给了我这个：

clang++ -o -std=c++11 test2.o -Wall -Werror                                                                                                                                                             ~
test2.o: In function `std::_Function_handler<opt<int> (int), main::$_0>::_M_invoke(std::_Any_data const&, int)':
test2.cc:(.text+0x1a7): undefined reference to `oopt<int>::~oopt()'
test2.o:(.rodata._ZTV4ooptIiE[_ZTV4ooptIiE]+0x10): undefined reference to `oopt<int>::~oopt()'
test2.o:(.rodata._ZTV4ooptIiE[_ZTV4ooptIiE]+0x18): undefined reference to `oopt<int>::~oopt()'
clang: error: linker command failed with exit code 1 (use -v to see invocati

上）

这不是一个精确的答案，但这可能会有很大帮助

score 0 · Accepted Answer

好吧，希望我的解决方案是正确的：至少它可以编译:)。我将把它作为答案，以便收到有关该解决方案的其他评论。再次感谢。

#include <iostream>
#include <unordered_map>
#include <functional>
#include <iostream>
#include <memory>

using namespace std;

//I need shared_pointers in order to gain the C++ dynamic casting
template <typename T> class Ptr {
    shared_ptr<T> ptr;

    public:
        Ptr(T* obj) : ptr{obj} {};
        Ptr(shared_ptr<T> obj) : ptr{obj} {};

        ~Ptr() {  };

        template <typename U> Ptr<U> Cast();

        T* operator->() const  {return &(*this->ptr);}
        operator T() const { return (*this->ptr); };

};

template <typename T> template <typename U> Ptr<U> Ptr<T>::Cast(){
    return Ptr<U>{dynamic_pointer_cast<U>(ptr)};
};


template <typename T> class opt {
    protected:
        T ret;                
    public:
       ~opt() {};
       virtual int isEmpty() = 0;
       virtual T* getObject() = 0;
       operator T() const { return ret; }; 
};

template <typename T> class oopt : public opt<T> {
    public:
    oopt(T obj) { 
        this->ret = obj; 
    };  
    int isEmpty() { return 1; };
    T* getObject() { return &this->ret; }
};

template <typename T> class eopt : public opt<T> {
    public:
    eopt() { };  //for downcasting matters
    T* getObject() { return nullptr; } //
    int isEmpty() { return 1; };
};

int main(void) {

    Ptr<oopt<int>> u(new oopt<int>(500));
    Ptr<opt<int>> w = u.Cast<opt<int>>();
    cout << (*(w->getObject())) << "\n" ;

    Ptr<opt<int>> none(new eopt<int>());
    w = none.Cast<opt<int>>();
    if ((w->getObject()) == nullptr) {
        cout << "null" << "\n";
    }

    return 1;

}

score 0 · Accepted Answer

好的，首先摆脱覆盖并使析构函数成为虚拟的（它们应该总是在继承时）。析构函数也未在基类中实现

#include <iostream>
#include <unordered_map>
#include <functional>
#include <iostream>
using namespace std;;

template <typename T> class opt {
    public:
        opt() {};
        virtual ~opt() {};
        virtual int isEmpty() = 0;
        virtual T getObject() = 0;
};

template <typename T> class oopt : public opt<T> {
    private:
        T ret;
    public:
        oopt(T obj) { ret = obj; };
        ~oopt(){};
        int isEmpty() { return 0; };
        T getObject() { return ret; };
};
int main(void) {

    function<opt<int>(int)> fu = [](int j) { return (oopt<int>(10)); };

    return 1;
}

现在编译。至于你最初得到的错误，我不能说它们来自哪里，因为我没有得到它们

c++ - C++ 中另一个带有闭包的抽象类问题

3 回答 3

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