我在创建基本文本用户界面时遇到问题。为了保持文本界面循环,我调用main(),直到用户选择按 0 退出。
但是,这会重新创建我的 LinkedList 列表,我想将其保留为永久列表。我知道将其作为全局变量是一种不好的做法,那么我该如何解决这个问题呢?
int main() {
int choice, newLatitude, newLongitude;
string newName;
LinkedList list;
cout << "[1] Add a city n";
cout << "[2] Display list of cities n";
cout << "[0] Exit program n";
cin >> choice;
if (choice == 0) {
return 0;
}
else if (choice == 1) {
cout << "Enter city name: ";
cin >> newName;
cout << "Enter latitude: ";
cin >> newLatitude;
cout << "Enter longitude: ";
cin >> newLongitude;
City newCity(newName, newLatitude, newLongitude);
list.addNode(newCity);
}
else if (choice == 2) {
list.display();
}
else {
cout << "Invalid option, please try again n";
}
main();
return 0;
}
使用递归来执行此任务可能不是一个好主意。如果你想让它重复,你可能最好有一个while循环。http://msdn.microsoft.com/en-us/library/2aeyhxcd.aspx
while(true) 只要语句 (true) 为真,就会循环。又名永远。你也可以做一些类似while(choice != 0)
的事情,尽管这需要对代码进行轻微的返工。
int main() {
int choice, newLatitude, newLongitude;
string newName;
LinkedList list;
while(true)
{
cout << "[1] Add a city n";
cout << "[2] Display list of cities n";
cout << "[0] Exit program n";
cin >> choice;
if (choice == 0) {
return 0;
}
else if (choice == 1) {
cout << "Enter city name: ";
cin >> newName;
cout << "Enter latitude: ";
cin >> newLatitude;
cout << "Enter longitude: ";
cin >> newLongitude;
City newCity(newName, newLatitude, newLongitude);
list.addNode(newCity);
}
else if (choice == 2) {
list.display();
}
else {
cout << "Invalid option, please try again n";
}
}
return 0;
}