Bra*_*ans 9 c++ enums parsing map
我已经将以下内容编写为文本冒险游戏的文本命令解析器的一部分.
我试图将用户输入的字符串与枚举类中的项相关联.以下是我的头文件:
#include <iostream>
#include <map>
#include <string>
using namespace std;
enum class Noun
{
// Interrogation subjects
name, // ask the subject his name
base, // where is the base?
attack, // when will the attack be?
invalid
};
map < string, Noun > knownNouns;
knownNouns["name"] = Noun::name;
knownNouns["base"] = Noun::base;
knownNouns["attack"] = Noun::attack;
Noun parseNoun(string &noun)
{
auto n = knownNouns.find(noun);
if ( n == knownNouns.end() ) {
return Noun::invalid;
}
return n->second;
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当我通过编译器,我得到以下内容:
nouns.h:46:1: error: 'knownNouns' does not name a type
knownNouns["name"] = Noun::name;
^
nouns.h:47:1: error: 'knownNouns' does not name a type
knownNouns["base"] = Noun::base;
^
nouns.h:48:1: error: 'knownNouns' does not name a type
knownNouns["attack"] = Noun::attack;
^
nouns.h: In function 'Noun parseNoun(std::string&)':
nouns.h:52:10: error: 'n' does not name a type
auto n = knownNouns.find(noun);
^
nouns.h:53:10: error: 'n' was not declared in this scope
if ( n == knownNouns.end() ) {
^
nouns.h:54:16: error: 'Noun' is not a class or namespace
return Noun::invalid;
^
nouns.h:56:12: error: 'n' was not declared in this scope
return n->second;
^
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这是我第一次尝试使用地图和枚举,而且我不确定我做错了什么.我对自动变速器的使用也不是非常熟悉,所以这对我来说是一些货物狂热编程.我希望我理解它的实现,一旦我解决了我遇到的类型定义问题,错误就会清除.
编辑:这很令人尴尬.我复制粘贴了一个我已经纠正过的错误.代码仍然无法编译时(即,同样的问题出现时)
map < string, Noun > knownNouns;
knownNouns["name"] = Verb::name;
knownNouns["base"] = Verb::base;
knownNouns["attack"] = Verb::attack;
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被纠正为
map < string, Noun > knownNouns;
knownNouns["name"] = Noun::name;
knownNouns["base"] = Noun::base;
knownNouns["attack"] = Noun::attack;
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Che*_*Alf 15
您不能将非声明构造直接放在命名空间范围中.
C++翻译单元是一系列声明.
分配等非声明语句必须在函数体内.
固定代码:
#include <iostream>
#include <map>
#include <string>
using namespace std;
enum class Noun
{
// Interrogation subjects
name, // ask the subject his name
base, // where is the base?
attack, // when will the attack be?
invalid
};
map< string, Noun > knownNouns =
{
{ "name", Noun::name },
{ "base", Noun::base },
{ "attack", Noun::attack }
};
auto parseNoun( string const& noun )
-> Noun
{
// auto n = knownNouns.find(noun);
// if ( n == knownNouns.end() ) {
// return Noun::invalid;
// }
// return n->second;
return Noun::invalid;
}
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这knownNouns是初始化的.这不是一项任务,即使=看起来非常像作业.
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