一种使用C ++ 0x枚举类中所有不合格名称的方法? [英] A way to use all the unqualified names in a C++0x enum class?
问题描述
枚举类E {
VAL1,VAL2
};
void fun(){
E e = E :: VAL1; //合格的名字
}
然而,我想知道如果我可以选择使用在一定范围内不合格的名称。如下:
void fun(){
using E :: *;
E e = VAL1;
switch(e){
case VAL2:...
我看到我可以使用E :: VAL1 编写并获取一个值。但是我不想为更大的枚举的每个值都这样做。
C ++ 11中没有办法做到这一点。为了防止这种情况你不知道 - 即使没有这个枚举,你也可以得到 E :: Val1
符号。对于这样的枚举,您可以使用 E ::
访问 Val1
。
但是,您不能采用范围枚举,并有选择地使其所有枚举器在给定范围内可见。还应该注意的是,您可以使用E :: Val1 编写 。该规范明确禁止这一点,您的编译器不会拒绝它。
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.
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.
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