C++:将枚举值打印为文本
如果我有这样的枚举
enum Errors
{ErrorA=0, ErrorB, ErrorC};
然后我想打印到控制台
Errors anError = ErrorA;
cout<<anError;/// 0 will be printed
但是我想要的是文本ErrorA",我可以在不使用 if/switch 的情况下做到吗?
您对此有何解决方案?
but what i want is the text "ErrorA", can i do it without using if/switch?
And what is your solution for this?
推荐答案
使用地图:
#include <iostream>
#include <map>
#include <string>
enum Errors {ErrorA=0, ErrorB, ErrorC};
std::ostream& operator<<(std::ostream& out, const Errors value){
static std::map<Errors, std::string> strings;
if (strings.size() == 0){
#define INSERT_ELEMENT(p) strings[p] = #p
INSERT_ELEMENT(ErrorA);
INSERT_ELEMENT(ErrorB);
INSERT_ELEMENT(ErrorC);
#undef INSERT_ELEMENT
}
return out << strings[value];
}
int main(int argc, char** argv){
std::cout << ErrorA << std::endl << ErrorB << std::endl << ErrorC << std::endl;
return 0;
}
在线性搜索中使用结构数组:
Using array of structures with linear search:
#include <iostream>
#include <string>
enum Errors {ErrorA=0, ErrorB, ErrorC};
std::ostream& operator<<(std::ostream& out, const Errors value){
#define MAPENTRY(p) {p, #p}
const struct MapEntry{
Errors value;
const char* str;
} entries[] = {
MAPENTRY(ErrorA),
MAPENTRY(ErrorB),
MAPENTRY(ErrorC),
{ErrorA, 0}//doesn't matter what is used instead of ErrorA here...
};
#undef MAPENTRY
const char* s = 0;
for (const MapEntry* i = entries; i->str; i++){
if (i->value == value){
s = i->str;
break;
}
}
return out << s;
}
int main(int argc, char** argv){
std::cout << ErrorA << std::endl << ErrorB << std::endl << ErrorC << std::endl;
return 0;
}
使用开关/案例:
#include <iostream>
#include <string>
enum Errors {ErrorA=0, ErrorB, ErrorC};
std::ostream& operator<<(std::ostream& out, const Errors value){
const char* s = 0;
#define PROCESS_VAL(p) case(p): s = #p; break;
switch(value){
PROCESS_VAL(ErrorA);
PROCESS_VAL(ErrorB);
PROCESS_VAL(ErrorC);
}
#undef PROCESS_VAL
return out << s;
}
int main(int argc, char** argv){
std::cout << ErrorA << std::endl << ErrorB << std::endl << ErrorC << std::endl;
return 0;
}
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