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Consider this one
template<typename T>
struct A {
enum class X {
V = T()
};
};
For member classes and member functions, C++11 (and C++03) won't instantiate their definition unless we use them in a way that requires their definition. Is this true for enum class?
// valid?
A<std::string> a;
Unfortunately, I can't check compilers, since C++11 is just out of the door and everything isn't reliable yet, it seems.
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You need to access toString() using any of these static instances or make the toString method static to have it accessible through CustomerType class. No, you cannot instantiate enums, and there's a good reason for that. Enums are for when you have a fixed set of related constants.
You can't create an instance of Enum using new operators. It should have a private constructor and is normally initialized as: ErrorCodes error = ErrorCodes. BUSSINESS_ERROR. Each constant in the enum has only one reference, which is created when it is first called or referenced in the code.
If the first enumerator has no initializer, the value of the corresponding constant is zero. An enumerator-definition without an initializer gives the enumerator the value obtained by increasing the value of the previous enumerator by one. So yes, if you do not specify a start value, it will default to 0.
The only difference is that enum constants are public , static and final (unchangeable - cannot be overridden). An enum cannot be used to create objects, and it cannot extend other classes (but it can implement interfaces).
I think so. 14.7.1/1
The implicit instantiation of a class template specialization causes the implicit instantiation of the declarations, but not of the definitions or default arguments, of [...] scoped member enumerations
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