我如何将枚举更改为字符？我正在尝试投射它，但它给了我一个错误

Question

所以我应该制定一个旅行计划来帮助旅行者了解他们在哪里（北、南、东、西）。我们应该为菜单选项创建枚举数据类型，而不是使用字符选项，用户选择他们要去的方向，它告诉总距离和离家的距离。 我尝试强制转换枚举，但出现此错误：预期的主要表达式之前；令牌。” 这是我的代码。

int main(){
    char option_letter;
    int lat1 = 0;
    int lat2 = 0;
    int long1 = 0;
    int long2 = 0;
    double distance = 0;
    int totalDistance = 0;
    enum direction_t {North = 'n', South = 's', East = 'e', West = 'w', Home = 'h', Quit = 'q'};
    direction_t direction = static_cast<direction_t>(tolower(option_letter));
    
    cout <<"Starting home at 0 North/South and 0 East/West, this program shows your position"<<endl;
    cout<<"after travelling 10 miles: north, south, east or west, how far you are from home"<<endl;
    cout<<"and how far you have traveled in total. You can jump home or quit." <<endl;
    
    distance = sqrt(pow(lat2-lat1, 2) + pow(long2-long1, 2));
    totalDistance = (lat1 + lat2) + (long1 + long2);
    while (true){
        cout <<"Location:"<< "longitude:" <<  long1<<" N"<< " latitude:" <<  lat1<<" E"<< " ""distance from home:"<< distance<< " ""distance traveled:"<< totalDistance<<endl;
        cout <<"Choose direction to travel(0 to quit) N)orth, S)outh, E)ast, W)est, H)ome, Q)uit:"<<endl;
        cin << direction_t;
        switch(direction_t){
            case 'n':{
                distance = sqrt(pow(lat2-lat1, 2) + pow(long2-long1, 2) * 1);
                totalDistance = (lat1+ lat2) + (long1 + long2);
                break;
            }
            case 'e':{
                distance = sqrt(pow(lat2-lat1, 2) + pow(long2-long1, 2) * 1);
                totalDistance = (lat1 + long1) + (long1 + long2);
                break;
            }
                
        }
        
    }
    return 0;
}

Answer 1

有几个问题：

首先cin 应该使用>> 而不是<<。

那么direction_t是一个类型，direction是一个变量。

最后，你不能直接读入direction_t 类型的变量，除非你为cin 重载>> 运算符。

Answer 2

" ""离家的距离："

你忘了删除额外的引号

除此之外，cin 需要右向箭头 (>>)

Answer 3

看起来转换实际上是正确实现的，但是，您从未初始化option_letter，因此您没有将任何数据传递给转换。在您的问题的上下文中，您应该通过cin >> option_letter 将数据从cin 传递给option_letter，然后将其转换为direction_t 类型。

您收到错误的原因是因为您在应该使用 direction 时使用了 direction_t（这发生在程序中的多个位置），并且正在使用 << 运算符和 @ 987654329@ 当您应该使用 >> 运算符时。

您的代码应如下所示--

#include <cctype>
#include <iostream>
#include <math.h>

using namespace std;

int main(){
    char option_letter;
    int lat1 = 0;
    int lat2 = 0;
    int long1 = 0;
    int long2 = 0;
    double distance = 0;
    int totalDistance = 0;
    enum direction_t {North = 'n', South = 's', East = 'e', West = 'w', Home = 'h', Quit = 'q'};
    //direction_t direction = static_cast<direction_t>(tolower(option_letter));
    
    cout <<"Starting home at 0 North/South and 0 East/West, this program shows your position"<<endl;
    cout<<"after travelling 10 miles: north, south, east or west, how far you are from home"<<endl;
    cout<<"and how far you have traveled in total. You can jump home or quit." <<endl;
    
    distance = sqrt(pow(lat2-lat1, 2) + pow(long2-long1, 2));
    totalDistance = (lat1 + lat2) + (long1 + long2);
    while (true){
        cout <<"Location:"<< "longitude:" <<  long1<<" N"<< " latitude:" <<  lat1<<" E"<< " ""distance from home:"<< distance<< " ""distance traveled:"<< totalDistance<<endl;
        cout <<"Choose direction to travel(0 to quit) N)orth, S)outh, E)ast, W)est, H)ome, Q)uit:"<<endl;
        cin >> option_letter;
        direction_t direction = static_cast<direction_t>(tolower(option_letter));

        switch(direction){
            case 'n':{
                distance = sqrt(pow(lat2-lat1, 2) + pow(long2-long1, 2) * 1);
                totalDistance = (lat1+ lat2) + (long1 + long2);
                break;
            }
            case 'e':{
                distance = sqrt(pow(lat2-lat1, 2) + pow(long2-long1, 2) * 1);
                totalDistance = (lat1 + long1) + (long1 + long2);
                break;
            }
                
        }
        
    }
    return 0;
}