Question

我和伙伴在我们的Text Adventure游戏中遇到了解析器问题。我们不断得到＆＃34;分段错误（核心转储）＆＃34;并且＆＃34;在抛出＆＃39; std :: out_of_range＆＃39;的实例后终止调用 what（）：basic_string :: substr 流产（核心倾销）＆＃34;错误，我们不知道热修复它。

 void nextTurn(Player *player)
{
    string input;
    Parser *parser;
    cout << player -> getName() << "'s turn: ";
    cin >> input;
    parser -> parse(input);

    delete parser;
}

这是我们从主课程中的玩家获得每个玩家每回合想要的内容的输入。这是我们的Parser.cpp classe。

  #include <iostream>
  #include "Parser.h"
  using namespace std;

  /**
   * Construct a new parser class
   */
  Parser::Parser(Player *p, TownArea *t)
  {
    Parser::player = p;
    Parser::townArea = t;
  }

 /**
 * Parse the input text
 */
  void Parser::parse(string input)
   {
            string action = input.substr(0, input.find(" "));

            //Look around the room
            if(action == "look" || action == "help")
            {
                    cout << "Current Room: " << Parser::townArea -> getName() << endl;
                    cout << Parser::townArea -> getDescription() << endl;
                    return;
            }

            //See player's inventory
            if(action == "inventory" || action == "inv")
            {
                    vector<string> inventory = Parser::player -> getInventory();

                    if(inventory.size() == 0) cout << "Inventory is empty." << endl;
                    else
                    {
                            cout << "Inventory: " << endl << "| ";
                            for(int i = 0; i < inventory.size(); i++)
                            {
                                    cout << inventory[i] << " | ";
                            }
                            cout << endl;
                    }
                    return;
            }
 }

Answer 1

Parser *parser;
cout << player -> getName() << "'s turn: ";
cin >> input;
parser -> parse(input);

您只是声明了Parser类的实例。您无法使用new构建它。未定义的行为。

Answer 2

您永远不会分配Parser：

void nextTurn(Player *player)
{
    string input;
    Parser *parser; // this points to nothing!!!!

    parser = new Parser; // now it points to a Parser

    cout << player -> getName() << "'s turn: ";
    cin >> input;
    parser -> parse(input);

    delete parser;
}

但是，当你在函数末尾只是delete时，就不需要使用指针。在这种情况下，只需使用正常的自动变量，如下所示：

void nextTurn(Player *player)
{
    string input;
    Parser parser; // Don't use a pointer!

    cout << player -> getName() << "'s turn: ";
    cin >> input;
    parser.parse(input);

    // no need to delete here
}

Player可能也是如此。只需创建它的正常实例并通过引用传递它：

void nextTurn(Player& player) // pass reference
{
    string input;
    Parser parser; // Don't use a pointer!

    cout << player.getName() << "'s turn: ";
    cin >> input;
    parser.parse(input);

    // no need to delete here
}

希望有所帮助。

Text Adventure Parser C ++

2 个答案: