c++的' operator function_type() '是可能的

C++ `operator function_type()` is such thing possible?

本文关键字：function operator c++ type      更新时间：2023-10-16

我尝试编译与如何在容器中存储具有不同签名的功能对象相关的答案(例如:std::map)我竞争(似乎对我来说)提供了答案代码:

#include <functional>
#include <iostream>
#include <string>
#include <map>
class api {
    //maps containing the different function pointers
    std::map<std::string, void(*)()> voida;
    std::map<std::string, int(*)(std::string, const int&)> stringcrint;
    friend class apitemp;
public:
    //api temp class 
    //given an api and a name, it converts to a function pointer  
    //depending on parameters used
    class apitemp {
        const std::string* n;
        api* p;
    public:
        apitemp(const std::string* name, const api* parent) 
            : n(name), p(parent) {}
        operator void(*)()() 
        {return p->void[*n];}
        operator int(*)(std::string, const int&)() 
        {return p->stringcrint[*n];}
    }; 
    //insertion of new functions into appropriate maps
    void insert(std::string name, void(*ptr)()) 
    {voida[name]=ptr;}
    void insert(std::string name, int(*ptr)(std::string, const int&))
    {stringcrint[name]=ptr;}
    //operator[] for the name gets halfway to the right function
    apitemp operator[](std::string n) const {return apitemp(n, this);}
} myMap;
int hello_world(std::string name, const int & number )
{
    name += "!";
    std::cout << "Hello, " << name << std::endl;
    return number;
}
int main() {
    myMap.insert("my_method_hello", &hello_world ); 
    //  int a = myMap["my_method_hello"]("Tim", 25);
}

但是我得到12个奇怪的错误在行与操作符:

Error 14  error C2665: 'api::apitemp::apitemp' : none of the 2 overloads could convert all the argument types 
Error 4   error C2586: incorrect user-defined conversion syntax : illegal indirections
Error 8   error C2586: incorrect user-defined conversion syntax : illegal indirections
Error 9   error C2440: 'initializing' : cannot convert from 'const api *' to 'api *'  
Error 10  error C2439: 'api::apitemp::p' : member could not be initialized    
Error 13  error C2232: '->api::stringcrint' : left operand has '' type, use '.'   
Error 2   error C2091: function returns function  
Error 3   error C2091: function returns function  
Error 6   error C2091: function returns function  
Error 7   error C2091: function returns function  
Error 11  error C2059: syntax error : '[' 
Error 1   error C2059: syntax error : '*'
Error 5   error C2059: syntax error : '*'
Error 12  error C2039: 'p' : is not a member of 'api'

所以我想知道-如何使它编译?

更新:修复后(感谢hvd的回答)我得到了这个:

#include <boost/function.hpp>
#include <iostream>
#include <string>
#include <map>
template <typename T> struct identity { typedef T type; };
class api {
    //maps containing the different function pointers
    std::map<std::string, identity<void(*)()>::type > voida;
    std::map<std::string, identity<int(*)(std::string, const int&)>::type > stringcrint;
    friend class apitemp;
public:
    //api temp class 
    //given an api and a name, it converts to a function pointer  
    //depending on parameters used
    class apitemp {
        std::string* n;
        api* p;
    public:
        apitemp(std::string* name, api* parent) 
            : n(name), p(parent) {}
        operator identity<void(*)()>::type()
        {return p->voida[*n];}
        operator identity<int(std::string, const int&)>::type*()
        {return p->stringcrint[*n];}
    }; 
    //insertion of new functions into appropriate maps
    void insert(std::string name, void(*ptr)()) 
    {voida[name]=ptr;}
    void insert(std::string name, int(*ptr)(std::string, const int&))
    {stringcrint[name]=ptr;}
    //operator[] for the name gets halfway to the right function
    apitemp operator[](std::string n) {return apitemp(n, this);}
} myMap;
int hello_world(std::string name, const int & number )
{
    name += "!";
    std::cout << "Hello, " << name << std::endl;
    return number;
}
int main() {
    myMap.insert("my_method_hello", &hello_world ); 
        int a = myMap["my_method_hello"]("Tim", 25);
}

仍然存在一个错误:

Error 1   error C2665: 'api::apitemp::apitemp' : none of the 2 overloads could convert all the argument types

可以使用函数指针类型的转换操作符，但语法不允许直接指定函数类型。您所需要做的就是使用一个typedef，我在这里把它包装在一个模板中:

template <typename T> struct identity { typedef T type; };
...
class api {
   // You can use identity<F*>::type
   operator identity<void(*)()>::type();
   // or you can use identity<F>::type*
   operator identity<int(std::string, const int&)>::type*();
};

该代码还有其他几个错误，例如使用const api*初始化api*，并在需要std::string*的地方传递std::string。

包含指向函数指针的声明令人费解，因此您可以先尝试使用typedef。

    typedef void (* no_arg_fun)();
    typedef int (* arg_fun)(std::string, const int&);
    operator no_arg_fun()
    {return p->voida[*n];}
    operator arg_fun()
    {return p->stringcrint[*n];}

你也有稳定性的问题。Map的operator[]是一个修改操作，所以你要么用map::find代替，要么把你自己的operator[]也改成非const。

也不清楚如何传递参数。例如，为什么传递const指针到字符串而不是const引用?为什么有些函数通过值传递字符串，而通过const引用传递整型(后者尤其没有意义，因为复制整型更便宜)。

既然我为你写了区域代码，我觉得有义务修复它:(

#include <functional>
#include <iostream>
#include <string>
#include <map>
class api {
    //maps containing the different function pointers
    typedef void(*voidfuncptr)();
    typedef int(*stringcrintptr)(std::string, const int&);
    std::map<std::string, voidfuncptr> voida;
    std::map<std::string, stringcrintptr> stringcrint;
public:
    //api temp class 
    //given an api and a name, it converts to a function pointer  
    //depending on parameters used
    class apitemp {
        const std::string n;
        const api* p;
    public:
        apitemp(const std::string& name, const api* parent) 
            : n(name), p(parent) {}
        operator voidfuncptr() 
        {return p->voida.find(n)->second;}
        operator stringcrintptr() 
        {return p->stringcrint.find(n)->second;}
    }; 
    //insertion of new functions into appropriate maps
    void insert(const std::string& name, voidfuncptr ptr) 
    {voida[name]=ptr;}
    void insert(const std::string& name, stringcrintptr ptr)
    {stringcrint[name]=ptr;}
    //operator[] for the name gets halfway to the right function
    apitemp operator[](std::string n) const {return apitemp(n, this);}
} myMap;
int hello_world(std::string name, const int & number )
{
    name += "!";
    std::cout << "Hello, " << name << std::endl;
    return number;
}
int main() {
    myMap.insert("my_method_hello", &hello_world ); 
    int a = myMap["my_method_hello"]("Tim", 25);
}
http://ideone.com/SXAPu

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