Enumerate over an enum in C++(在 C++ 中枚举枚举)
问题描述
在 C++ 中,是否可以枚举枚举(运行时或编译时(首选))并为每次迭代调用函数/生成代码?
In C++, Is it possible to enumerate over an enum (either runtime or compile time (preferred)) and call functions/generate code for each iteration?
示例用例:
enum abc
{
start
a,
b,
c,
end
}
for each (__enum__member__ in abc)
{
function_call(__enum__member__);
}
<小时>
似是而非的重复:
Plausible duplicates:
- C++:遍历枚举
- C++ 中的 Enum 就像 Ada 中的 Enum?
推荐答案
要添加到 @StackedCrooked 答案,您可以重载 operator++
, operator--
和 operator*
并具有类似迭代器的功能.
To add to @StackedCrooked answer, you can overload operator++
, operator--
and operator*
and have iterator like functionality.
enum Color {
Color_Begin,
Color_Red = Color_Begin,
Color_Orange,
Color_Yellow,
Color_Green,
Color_Blue,
Color_Indigo,
Color_Violet,
Color_End
};
namespace std {
template<>
struct iterator_traits<Color> {
typedef Color value_type;
typedef int difference_type;
typedef Color *pointer;
typedef Color &reference;
typedef std::bidirectional_iterator_tag
iterator_category;
};
}
Color &operator++(Color &c) {
assert(c != Color_End);
c = static_cast<Color>(c + 1);
return c;
}
Color operator++(Color &c, int) {
assert(c != Color_End);
++c;
return static_cast<Color>(c - 1);
}
Color &operator--(Color &c) {
assert(c != Color_Begin);
return c = static_cast<Color>(c - 1);
}
Color operator--(Color &c, int) {
assert(c != Color_Begin);
--c;
return static_cast<Color>(c + 1);
}
Color operator*(Color c) {
assert(c != Color_End);
return c;
}
让我们用一些模板进行测试
Let's test with some <algorithm>
template
void print(Color c) {
std::cout << c << std::endl;
}
int main() {
std::for_each(Color_Begin, Color_End, &print);
}
现在,Color
是一个常量双向迭代器.这是我在上面手动执行时编码的可重用类.我注意到它可以用于更多的枚举,所以一遍又一遍地重复相同的代码是很乏味的
Now, Color
is a constant bidirectional iterator. Here is a reusable class i coded while doing it manually above. I noticed it could work for many more enums, so repeating the same code all over again is quite tedious
// Code for testing enum_iterator
// --------------------------------
namespace color_test {
enum Color {
Color_Begin,
Color_Red = Color_Begin,
Color_Orange,
Color_Yellow,
Color_Green,
Color_Blue,
Color_Indigo,
Color_Violet,
Color_End
};
Color begin(enum_identity<Color>) {
return Color_Begin;
}
Color end(enum_identity<Color>) {
return Color_End;
}
}
void print(color_test::Color c) {
std::cout << c << std::endl;
}
int main() {
enum_iterator<color_test::Color> b = color_test::Color_Begin, e;
while(b != e)
print(*b++);
}
实施如下.
template<typename T>
struct enum_identity {
typedef T type;
};
namespace details {
void begin();
void end();
}
template<typename Enum>
struct enum_iterator
: std::iterator<std::bidirectional_iterator_tag,
Enum> {
enum_iterator():c(end()) { }
enum_iterator(Enum c):c(c) {
assert(c >= begin() && c <= end());
}
enum_iterator &operator=(Enum c) {
assert(c >= begin() && c <= end());
this->c = c;
return *this;
}
static Enum begin() {
using details::begin; // re-enable ADL
return begin(enum_identity<Enum>());
}
static Enum end() {
using details::end; // re-enable ADL
return end(enum_identity<Enum>());
}
enum_iterator &operator++() {
assert(c != end() && "incrementing past end?");
c = static_cast<Enum>(c + 1);
return *this;
}
enum_iterator operator++(int) {
assert(c != end() && "incrementing past end?");
enum_iterator cpy(*this);
++*this;
return cpy;
}
enum_iterator &operator--() {
assert(c != begin() && "decrementing beyond begin?");
c = static_cast<Enum>(c - 1);
return *this;
}
enum_iterator operator--(int) {
assert(c != begin() && "decrementing beyond begin?");
enum_iterator cpy(*this);
--*this;
return cpy;
}
Enum operator*() {
assert(c != end() && "cannot dereference end iterator");
return c;
}
Enum get_enum() const {
return c;
}
private:
Enum c;
};
template<typename Enum>
bool operator==(enum_iterator<Enum> e1, enum_iterator<Enum> e2) {
return e1.get_enum() == e2.get_enum();
}
template<typename Enum>
bool operator!=(enum_iterator<Enum> e1, enum_iterator<Enum> e2) {
return !(e1 == e2);
}
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本文标题为:在 C++ 中枚举枚举


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