C ++函数图

Question

我如何调用这些函数？

假设我有3个显示功能

void disp1()
{
  std::cout<<"\n Disp1 ";
}

void disp2()
{
  std::cout<<"\n Disp2 ";
}

void disp3()
{
  std::cout<<"\n Disp3 ";
}

我想创建这些函数的映射，并根据键调用它们。

fMap["d1"] 

应运行该disp1函数。我将如何解决这个问题;

如果函数的返回类型不是void，我可以获取值但是我需要打印的语句怎么样。像上面那些;

Answer 1

您可以使用std::map和std::function执行此操作：

#include <map>
#include <string>
#include <iostream>
#include <functional>

void disp1()
{
  std::cout<<"\n Disp1 ";
}

void disp2()
{
  std::cout<<"\n Disp2 ";
}

void disp3()
{
  std::cout<<"\n Disp3 ";
}

int main()
{
    // add the functions as an initializer-list
    std::map<std::string, std::function<void()>> fMap
    {
          {"d1", disp1}
        , {"d2", disp2}
        , {"d3", disp3}
    };

    // or add them one at a time
    fMap["d1"] = disp1;
    fMap["d2"] = disp2;
    fMap["d3"] = disp3;

    fMap["d1"](); // call them using ()
    fMap["d2"]();
    fMap["d3"]();
}

Answer 2

要存储查找索引，您可以使用std::map。这包括键值对（mymap[ key ] = value）。 C ++允许您存储指向函数的指针，在这里我使用using关键字创建一个类型名称func_t，它的函数将具有指纹类型：void (*)(void)（这就是你说的＆＃34;指向一个取消void并返回void的函数的指针。

#include <iostream>
#include <string>
#include <map>

// the functions we intend to map
void disp1()
{
  std::cout<<"Disp1\n";
}

void disp2()
{
  std::cout<<"Disp2\n";
}

void disp3()
{
  std::cout<<"Disp3\n";
}

int main() {
    // create a new type, func_t, which describes a pointer
    // to a void function that takes no parameters (void).
    using func_t = void (*)(void);
    // declare a map from a string key to a func_t value,
    // and initialize it with a mapping of f1->disp1, f2->disp2
    // and f3->disp3
    std::map<std::string, func_t> functionMap = {
        { "f1", disp1 }, { "f2", disp2 }, { "f3", disp3 }
    };

    // declare a string for reading input
    std::string input;

    // loop until there is no more input on std::cin.
    while (std::cin.good()) {
        // prompt
        std::cout << "Which disp (f1, f2, f3)? ";
        // fetch the next line of text from cin, without the \n
        std::getline(std::cin, input);
        // if the input is empty we ran out of input or the user
        // input a blank line. either way, stop.
        if (input.empty())
            break;

        std::cout << "You chose " << input << "\n";

        // look for the key in the map. if the key is not found,
        // it will equal the special iterator functionMap.end()
        auto it = functionMap.find(input);
        // If it's not functionMap.end then we have a valid iterator.
        // it->first is the key, it->second is the function pointer.
        if (it != functionMap.end()) {
            // to use a function pointer, just add the () with any
            // arguments after the variable name.
            // remember, it->second is the function pointer.
            it->second();
        } else {
            std::cout << "Invalid entry.\n";
        }
    }
}

现场演示：http://ideone.com/4Xlow1

Answer 3

第一个问题的答案是在地图中使用std::function。

    std::map<std::string, std::function<void>> fMap {
        {"d1", disp1},
        {"d2", disp2},
        {"d3", disp3}
    };

关于我需要考虑的返回值的第二个问题。

Answer 4

在@ Galik的评论

的帮助下解决

   #include <iostream>
#include <functional>
#include <algorithm>
#include <map>



typedef std::function<void()> func_t;
typedef std::map<std::string, func_t> func_t_map;

void disp1()
{
    std::cout<<"\n Display 1 "<<std::endl;
    return;
}


void disp2()
{
    std::cout<<"\n Display 2 "<<std::endl;
    return;
}

void disp3()
{
    std::cout<<"\n Display 3 "<<std::endl;
    return;
}

void disp4()
{
    std::cout<<"\n Display 4 "<<std::endl;
    return;
}



int main()
{
    func_t_map fMap;


    fMap["d1"] = disp1;
    fMap["d2"] = disp2;
    fMap["d3"] = disp3;
    fMap["d4"] = disp4;

    fMap["d2"]();

    return 0;
}