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
的(模板)函数,它能够执行如下操作:
OPTION( CURSOR::ON );
OPTION( ANSI::ON );
我试过这样实现:
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...;
}
但是如果我尝试编译之前定义的两行代码,它会给我以下错误:
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();
请忽略我在代码中的命名空间 osm
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。
最佳答案
一个选择是编写多个重载函数,每个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
https://stackoverflow.com/questions/72746726/