以不同枚举类类型作为输入的函数，如何？

Question

我面临以下问题。假设我有两个（或更多）这样的枚举类：

enum class CURSOR { ON, OFF };
enum class ANSI { ON, OFF };

我正在尝试实现一个名为“OPTION能够执行以下操作”的（模板）函数：

OPTION( CURSOR::ON );
OPTION( ANSI::ON );

我尝试以这种方式实现它：

template <typename T>
inline void OPTION( const T& opt )
 {
  if( opt == CURSOR::ON ) //do something...;
  else if( opt == CURSOR::OFF ) //do something...; 
  else if( opt == ANSI::ON ) //do something...;
  else if( opt == ANSI::OFF ) //do something...;
 }

但如果我尝试编译之前定义的两行代码，则会出现以下错误：

examples/../include/manipulators/csmanip.hpp: In instantiation of 'void osm::OPTION(const T&) [with T = CURSOR]':
examples/manipulators.cpp:190:14:   required from here
examples/../include/manipulators/csmanip.hpp:88:18: error: no match for 'operator==' (operand types are 'const osm::CURSOR' and 'osm::ANSI')
   88 |     else if( opt == ANSI::ON ) enableANSI();
      |              ~~~~^~~~~~~~~~~
examples/../include/manipulators/csmanip.hpp:88:18: note: candidate: 'operator==(osm::ANSI, osm::ANSI)' (built-in)
examples/../include/manipulators/csmanip.hpp:88:18: note:   no known conversion for argument 1 from 'const osm::CURSOR' to 'osm::ANSI'
examples/../include/manipulators/csmanip.hpp:88:18: note: candidate: 'operator==(osm::CURSOR, osm::CURSOR)' (built-in)
examples/../include/manipulators/csmanip.hpp:88:18: note:   no known conversion for argument 2 from 'osm::ANSI' to 'osm::CURSOR'
examples/../include/manipulators/csmanip.hpp:89:18: error: no match for 'operator==' (operand types are 'const osm::CURSOR' and 'osm::ANSI')
   89 |     else if( opt == ANSI::OFF ) disableANSI();

osm请忽略我在代码中的命名空间中定义它们的事实。

您知道问题是什么吗？您是否知道如何构建适用于此任务的函数？谢谢。

Answer 1

一种选择是编写多个重载函数，每个重载函数处理不同的枚举类型。例如：

void OPTION(CURSOR c) {
    switch (c) {
        case CURSOR::ON:  /* turn cursor on */;  break;
        case CURSOR::OFF: /* turn cursor off */; break;
    }
}

void OPTION(ANSI a) {
    switch (a) {
        case ANSI::ON:  /* turn ANSI on */;  break;
        case ANSI::OFF: /* turn ANSI off */; break;
    }
}

然后重载解析将选择正确的函数来调用：

OPTION(CURSOR::OFF); // Calls first function
OPTION(ANSI::ON);    // Calls second function   

Answer 2

我同意 273K 和 templatetypedef 关于使用重载的观点。

但是，如果您希望拥有一个将案例处理逻辑全部集中在一个地方的模板，您可以这样做：

#include <iostream>
#include <type_traits>

using std::cout;

enum class CURSOR { ON, OFF };
enum class ANSI { ON, OFF };

template <typename T>
void OPTION(T opt) {
    if constexpr (std::is_same_v<T, CURSOR>) {
        if (opt == CURSOR::ON) cout << "cursor is on\n";
        else if (opt == CURSOR::OFF) cout << "cursor is off\n";
    } else if constexpr (std::is_same_v<T, ANSI>) {
        if (opt == ANSI::ON) cout << "ANSI is on\n";
        else if (opt == ANSI::OFF) cout << "ANSI is off\n";
    }
}

int main() {
    OPTION( CURSOR::ON );
    OPTION( ANSI::ON );
}