Tia*_*ago 19 c c++ macros variadic c-preprocessor
为了使我的代码更短更容易更改,我想替换类似的东西
enum{ E_AAA, E_BBB, E_CCC };
static const char *strings{"AAA", "BBB", "CCC" };
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使用宏,如INIT(AAA,BBB,CCC); 但是当我尝试使用变量参数和字符串化做一个宏时,我得到一个错误,因为没有声明参数.
有关如何做到这一点的任何想法?
Dr *_*eco 27
这是我几天前学到的解决方案.参加您的问题的简化版本是:
#define ENUM_MACRO(name, v1, v2, v3, v4, v5, v6, v7)\
enum name { v1, v2, v3, v4, v5, v6, v7};\
const char *name##Strings[] = { #v1, #v2, #v3, #v4, #v5, #v6, #v7};
ENUM_MACRO(Week, Sun, Mon, Tue, Wed, Thu, Fri, Sat);
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但是你可以使用函数调用来改进版本,如下所示:
#define ENUM_MACRO(name, v1, v2, v3, v4, v5, v6, v7)\
enum name { v1, v2, v3, v4, v5, v6, v7};\
const char *name##Strings[] = { #v1, #v2, #v3, #v4, #v5, #v6, #v7};\
const char *name##ToString(value) { return name##Strings[value]; }
ENUM_MACRO(Week, Sun, Mon, Tue, Wed, Thu, Fri, Sat);
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这将成长为:
enum Week { Sun, Mon, Tue, Wed, Thu, Fri, Sat};
const char *WeekStrings[] = { "Sun", "Mon", "Tue", "Wed", "Thu", "Fri", "Sat"};
const char *WeekToString(value) { return WeekStrings[value]; };
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您甚至可以为第一个元素使用偏移量,如下所示:
#define ENUM_MACRO(name, offset, v1, v2, v3, v4, v5, v6, v7)\
enum name { v1 = offset, v2, v3, v4, v5, v6, v7};\
const char *name##Strings[] = { #v1, #v2, #v3, #v4, #v5, #v6, #v7};\
const char *name##ToString(value) { return name##Strings[value - offset ]; }
ENUM_MACRO(Week, 1, Sun, Mon, Tue, Wed, Thu, Fri, Sat);
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我希望这有帮助.
小心,Beco
参考:
打印月份问题,由Kush,Danny Varod回答
Sco*_*ley 14
你可以用一些宏魔法来做到这一点:
#define FRUITS \
etype(Unknown), \
etype(Apple), \
etype(Orange), \
etype(Banana), \
etype(Apricot), \
etype(Mango)
#define etype(x) F_##x
typedef enum { FRUITS } Fruit;
#undef etype
#define etype(x) #x
static const char *strFruit[] = { FRUITS };
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这是一个测试程序:
#include <iostream>
#include <exception>
#include <vector>
#define FRUITS \
etype(Unknown), \
etype(Apple), \
etype(Orange), \
etype(Banana), \
etype(Apricot), \
etype(Mango)
#define etype(x) F_##x
typedef enum { FRUITS } Fruit;
#undef etype
#define etype(x) #x
static const char *strFruit[] = { FRUITS };
const char *enum2str (Fruit f)
{
return strFruit[static_cast<int>(f)];
}
Fruit str2enum (const char *f)
{
const int n = sizeof(strFruit) / sizeof(strFruit[0]);
for (int i = 0; i < n; ++i)
{
if (strcmp(strFruit[i], f) == 0)
return (Fruit) i;
}
return F_Unknown;
}
int main (int argc, char *argv[])
{
std::cout << "I like " << enum2str(F_Mango) << std::endl;
std::cout << "I do not like " << enum2str(F_Banana) << std::endl;
std::vector<char *> v;
v.push_back("Apple");
v.push_back("Mango");
v.push_back("Tomato");
for (int i = 0; i < v.size(); ++i)
{
const Fruit f = str2enum(v[i]);
if (f == F_Unknown)
std::cout << "Is " << v[i] << " a fruit?" << std::endl;
else
std::cout << v[i] << " is a fruit" << std::endl;
}
return 0;
}
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它输出:
I like Mango
I do not like Banana
Apple is a fruit
Mango is a fruit
Is Tomato a fruit?
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小智 11
这是我的解决方案:
#define FRUITS(fruit) \
fruit(Apple) \
fruit(Orange) \
fruit(Banana)
#define CREATE_ENUM(name) \
F_##name,
#define CREATE_STRINGS(name) \
#name,
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诀窍在于,"水果"是宏"水果"的一个参数,将被你传递给它的东西取代.例如:
FRUITS(CREATE_ENUM)
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将扩展到这个:
F_Apple, F_Orange, F_Banana,
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让我们创建枚举和字符串数组:
enum fruit {
FRUITS(CREATE_ENUM)
};
const char* fruit_names[] = {
FRUITS(CREATE_STRINGS)
};
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这是Boost.Preprocessor的解决方案:
#include <boost/preprocessor.hpp>
#define DEFINE_ENUM_DECL_VAL(r, name, val) BOOST_PP_CAT(name, BOOST_PP_CAT(_, val))
#define DEFINE_ENUM_VAL_STR(r, name, val) BOOST_PP_STRINGIZE(val)
#define DEFINE_ENUM(name, val_seq) \
enum name { \
BOOST_PP_SEQ_ENUM(BOOST_PP_SEQ_TRANSFORM(DEFINE_ENUM_DECL_VAL, name, val_seq)) \
}; \
static const char* BOOST_PP_CAT(name, _strings[] = ) { \
BOOST_PP_SEQ_ENUM(BOOST_PP_SEQ_TRANSFORM(DEFINE_ENUM_VAL_STR, name, val_seq)) \
};
DEFINE_ENUM(E, (AAA)(BBB)(CCC))
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(AAA)(BBB)(CCC)是树元素AAA,BBB和CCC的Boost.Preprocessor序列;宏将枚举名称附加到其模态中:
enum E { E_AAA, E_BBB, E_CCC };
static const char* E_strings[] = { "AAA", "BBB", "CCC" };
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