枚举在C ++中的枚举 [英] Enumerate over an enum in C++
问题描述
在C ++中,是否可以枚举一个枚举(运行时或编译时(首选)),并调用函数/生成每个迭代的代码?
示例用例:
枚举abc
{
开始
a,
b,
c,
end
}
每个(__enum__member__在abc)
{
function_call(__ enum__member__);
}
合理的重复: p>
要添加到 @StackedCrooked 回答,你可以重载 operator ++
, operator -
和 operator *
并具有类似功能的迭代器。
枚举颜色{
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;
};
}
颜色&运算符++(颜色& c){
assert(c!= Color_End);
c = static_cast< Color>(c + 1);
return c;
}
颜色运算符++(Color& c,int){
assert(c!= Color_End);
++ c;
return static_cast< Color>(c - 1);
}
颜色&运算符 - (Color& c){
assert(c!= Color_Begin);
return c = static_cast< Color>(c - 1);
}
颜色运算符 - (Color& c,int){
assert(c!= Color_Begin);
--c;
return static_cast< Color>(c + 1);
}
颜色运算符*(颜色c){
assert(c!= Color_End);
return c;
}
让我们用一些< algorithm>
模板
void print(Color c){
std :: cout< c <的std :: ENDL;
}
int main(){
std :: for_each(Color_Begin,Color_End,& print);
}
现在,颜色
是一个恒定的双向迭代器。这是一个可重用的类,我在上面手动进行编码。我注意到它可以工作更多的枚举,所以重复相同的代码一遍又一遍是相当乏味的
//测试代码enum_iterator
// --------------------------------
命名空间color_test {
枚举颜色{
Color_Begin,
Color_Red = Color_Begin,
Color_Orange,
Color_Yellow,
Color_Green,
Color_Blue,
Color_Indigo ,
Color_Violet,
Color_End
};
颜色begin(enum_identity< Color>){
return Color_Begin;
}
颜色结束(enum_identity< Color>){
return Color_End;
}
}
void print(color_test :: Color c){
std :: cout<<< c <的std :: ENDL;
}
int main(){
enum_iterator< color_test :: Color> b = color_test :: Color_Begin,e;
while(b!= e)
print(* b ++);
}
执行如下。
模板< typename T>
struct enum_identity {
typedef T type;
};
命名空间详细信息{
void begin();
void end();
}
模板< typename枚举>
struct enum_iterator
:std :: iterator< std :: bidirectional_iterator_tag,
枚举> {
enum_iterator():c(end()){}
enum_iterator(枚举c):c(c){
assert(c> = begin() ;& c< = end());
}
enum_iterator& operator =(Enum c){
assert(c> = begin()&& c< = end());
this-> c = c;
return * this;
}
static枚举begin(){
使用细节:: begin; // re-enable ADL
return begin(enum_identity< Enum>());
}
static枚举结束(){
使用细节:: end; //重新启用ADL
return end(enum_identity< Enum>());
}
enum_iterator& operator ++(){
assert(c!= end()&&增量过去结束?
c = static_cast&Enum>(c + 1);
return * this;
}
enum_iterator操作符++(int){
assert(c!= end()&&增量过去结束?
enum_iterator cpy(* this);
++ * this;
return cpy;
}
enum_iterator& operator - (){
assert(c!= begin()& amp;&减量超出开始?
c = static_cast<枚举>(c - 1);
return * this;
}
enum_iterator运算符 - (int){
assert(c!= begin()&&减量超出开始?
enum_iterator cpy(* this);
- * this;
return cpy;
}
枚举算子*(){
assert(c!= end()&&不能取消引用结束迭代器);
return c;
}
枚举get_enum()const {
return c;
}
private:
枚举c;
};
template< typename枚举>
bool operator ==(enum_iterator< Enum> e1,enum_iterator&Enum> e2){
return e1.get_enum()== e2.get_enum();
}
模板< typename枚举>
bool operator!=(enum_iterator&Enum> e1,enum_iterator&Enum> e2){
return!(e1 == e2);
}
In C++, Is it possible to enumerate over an enum (either runtime or compile time (preferred)) and call functions/generate code for each iteration?
Sample use case:
enum abc
{
start
a,
b,
c,
end
}
for each (__enum__member__ in abc)
{
function_call(__enum__member__);
}
Plausible duplicates:
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;
}
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);
}
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++);
}
Implementation follows.
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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