我有一个具有枚举值作为参数的类。它具有m_ConsistencyErrors
成员std::set
。我想根据枚举参数的值将该成员的类型设置为构造。
如果TestType值为
类型MSG123_CONSISTENCY_TEST
,我希望M_ConsistencyErrors为std::set<EnMsg123Param>
如果TestType值为
MSG5_CONSISTENCY_TEST
,我希望M_ConsistencyErrors为std::set<EnMsg5Param>
有一种干净的方法可以实现这一目标,或者我应该找到另一种解决方案。
class CMsgConsistencyTest // : public CTestBase // left out for simplicity
{
enum EnTests
{
MSG123_CONSISTENCY_TEST,
MSG5_CONSISTENCY_TEST,
};
enum EnMsg123Param
{
Msg123_1,
Msg123_2
};
enum EnMsg5Param
{
Msg5_1,
Msg5_2
};
public:
CMsgConsistencyTest(const EnTests TestType) // : CTestBase(TestType) // left out for simplicity
{
if (TestType == MSG123_CONSISTENCY_TEST)
{
ParameterType = EnMsg123Param; // pseudo code
}
else if (TestType == MSG5_CONSISTENCY_TEST)
{
ParameterType = EnMsg5Param; // pseudo code
}
}
private:
template<typename ParameterType>
std::set<ParameterType> m_ConsistencyErrors;
};
您无法执行此操作, parametertype 必须在使用 cmsgconsistencyTest 以及其成员访问m_ConsistencyErrors
对于该 cmsgconsistencyTest 可以是模板类,示例
#include <set>
enum EnMsg123Param
{
Msg123_1,
Msg123_2,
};
enum EnMsg5Param
{
Msg5_1,
Msg5_2,
Msg5_3,
};
template<typename ParameterType>
class CMsgConsistencyTest // : public CTestBase // left out for simplicity
{
public:
// ...
private:
std::set<ParameterType> m_ConsistencyErrors;
};
// and for instance
CMsgConsistencyTest<EnMsg123Param> A;
CMsgConsistencyTest<EnMsg5Param> B;
否则,您可能必须做一些丑陋,灾难性的和"非C "的事情:
CMsgConsistencyTest(const EnTests TestType) // : CTestBase(TestType) // left out for simplicity
{
if (TestType == MSG123_CONSISTENCY_TEST)
{
m_ConsistencyErrors = new set<EnMsg123Param>;
}
else if (TestType == MSG5_CONSISTENCY_TEST)
{
m_ConsistencyErrors = new set<EnMsg5Param>;
}
// else ?
// probably need to save TestType etc
}
private:
void * m_ConsistencyErrors;