我如何將枚舉更改為字符？ 我正在嘗試投射它，但它給了我一個錯誤

Question

所以我應該制定一個旅行計划來幫助旅行者知道他們在哪里（北，南，東，西）。 我們應該為菜單選項創建枚舉數據類型，而不是使用字符選項，用戶選擇他們要去的方向，它告訴總距離和離家的距離。 我嘗試強制轉換枚舉，但出現此錯誤：預期的主表達式之前； 令牌。”這是我的代碼。

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

" ""離家距離："

你忘了刪除額外的引號

除此之外，您需要 cin 的右向箭頭 (>>)

Answer 2

有幾個問題：

首先cin應該使用>>而不是<< 。

那么direction_t是一個類型， direction是一個變量。

最后你不能直接讀入direction_t類型的變量，除非你重載cin 的>>操作符。

Answer 3

看起來強制轉換實際上是正確實現的，但是，您從未初始化option_letter ，因此您沒有將任何數據傳遞給強制轉換。 在您的問題的上下文中，您應該通過cin >> option_letter將數據從cin傳遞到option_letter ，然后將其轉換為direction_t類型。

您收到錯誤的原因是因為您在應該使用direction_t時使用了direction （這發生在程序中的多個位置），並且在您應該使用>>時將<<運算符與cin一起使用操作員。

你的代碼應該是這樣的——

#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;
}