问题
The new C++ (C++0x or C++11) has an new kind of enum, an "enum class" where the names are scoped to the enum (among other things).
enum class E {
VAL1, VAL2
};
void fun() {
E e = E::VAL1; // Qualified name
}
I'm wondering, however, if I can selectively use the unqualified name in a certain scope. Something like:
void fun() {
using E::*;
E e = VAL1;
switch (e) {
case VAL2: ...
I see I can write using E::VAL1
and get one value. But I don't want to do that for every value of a larger enum.
回答1:
There is no way to do this in C++11. Just in case you are not aware of it - you get the E::Val1
notation even for an unscoped enumeration. For such an enumeration, you have Val1
accessible with and without the use of E::
.
But you cannot take a scoped enumeration and selectively make all its enumerators visible in a given scope. It should also be noted that you can not write using E::Val1
. The spec explicitly forbids this, your compiler just doesn't reject it yet.
回答2:
This is also something I happen to want, but haven't gotten around to trying to solve. Here's an untested solution. EDIT: I tried it out and it works great! This is my very first C++11 utility macro. Also I added a one-past-the-end enumerator to help extend it to "derived" enumerations.
#define IMPORTABLE_ENUM( TYPENAME, ... ) \
\
struct import_ ## TYPENAME { \
enum TYPENAME { \
__VA_ARGS__ \
}; \
}; \
\
typedef import_ ## TYPENAME :: TYPENAME TYPENAME;
This cannot be imported into a block scope, but defines a base class import_duck
to bring the enumerators into a class. Usage:
IMPORTABLE_ENUM ( duck, huey, dewey, louie )
duck d = duck::dewey; // can't use unscoped enumerators here
struct duck_madness : private import_duck { // but inside a derived class
duck who_did_it() { return huey; } // qualification is unnecessary
}
And since there is only one enum
type even among derived classes, no static_cast
is ever required.
来源:https://stackoverflow.com/questions/7457058/a-way-to-use-all-the-unqualified-names-in-a-c0x-enum-class