c++ - 当程序预期根据用户输入再次循环时，它反而会导致段错误 : 11

Question

我用 C++ 编写了一个简单的剪刀石头布程序，并尝试以这样一种方式实现 main.cpp 文件:玩完游戏后，如果用户想再次玩游戏，系统会要求用户输入。 'y' 的响应应该再次运行程序，允许用户输入另一个名称和移动选择，但是程序只会成功运行一次，然后导致:运行完成:段错误:11。

代码:

主要.cpp

#include <iostream>
#include <string>
#include "rock-paper-scissorss.h"
using namespace std;

void whoWon(string player_move, string computer_move){
    if(player_move == "rock"){
        if(computer_move == "paper")
            cout<<"You chose rock and the computer chose paper, the computer wins!"<<endl;

        else if(computer_move == "scissors")
            cout<<"You chose rock and the computer chose scissors, you win!"<<endl;

        else
            cout<<"You chose rock and the computer chose rock, it's a tie!"<<endl;
    }

    else if(player_move == "paper"){
        if(computer_move == "scissors")
            cout<<"You chose paper and the computer chose scissors, the computer wins!"<<endl;

        else if(computer_move == "rock")
            cout<<"You chose paper and the computer chose rock, you win!"<<endl;

        else
            cout<< "You chose paper and the computer chose paper, it's a tie!"<<endl;
    }

    else{
        if(computer_move == "rock")
            cout<<"You chose scissors and the computer chose rock, the computer wins!"<<endl;

        else if(computer_move == "paper")
            cout<<"You chose scissors and the computer chose paper, you win!"<<endl;

        else
            cout<<"You chose scissors and the computer chose scissors, it's a tie!"<<endl;
    }
}

int main(int argc, char** argv) {

  char play_again = 'y';

  while(play_again == 'y'){
    cout<<"Please enter your name: "<<endl;
    string name;
    getline(cin,name);
    cout<<"Hello "<<name<<", welcome to Rock-Paper-Scissors!"<<endl;

    Player player_1(name);

    cout<<"Please enter the move you wish to make: "<<endl;
    string move;
    getline(cin,move);
    while(1){
        if((move == "rock") || (move == "paper") || (move == "scissors")){

            player_1.setMoveChoice(move);
            break;
        }

        else{
            cout<<"Invalid move choice! Please enter rock, paper, or scissors"<<endl;
            getline(cin,move);
        }
    } 


    Computer com_1;
    com_1.setMoveChoice();

   string p_move = player_1.getMoveChoice();
   string c_move = com_1.getMoveChoice();


   whoWon(p_move,c_move);



   cout<<"Would you like to play again?(y/n)"<<endl;

   cin>>play_again;
   cout<<"your response was: "<<play_again<<endl;



    } 



 return 0;
}

剪刀石头布.cpp:

#include "rock-paper-scissorss.h"
#include <cstdlib>

Computer :: Computer(){
    num_generated = rand() % 3 + 1;
}

void Computer ::  setMoveChoice(){
   if(num_generated == 1)
       move_selected = "rock";

   else if(num_generated == 2)
       move_selected = "paper";

   else
       move_selected = "scissors";

}

string Computer :: getMoveChoice(){
    return move_selected;
}

Computer :: ~Computer(){

    delete this;
}


Player :: Player(string name){
    player_name = name;

}

void Player :: setMoveChoice(string choice){
    move_selected = choice;
}

string Player :: getMoveChoice(){
    return move_selected;
}

Player :: ~Player(){
    delete this;
}

剪刀石头布.h:

#ifndef ROCK_PAPER_SCISSORSS_H
#define ROCK_PAPER_SCISSORSS_H

#include <iostream>
#include <string>
using namespace std;



class Computer{

private:
    int num_generated;
    string move_selected;

public:
    Computer();
    void setMoveChoice();
    string getMoveChoice();
    ~Computer();
};

class Player{

private:
    string move_selected;
    string player_name;

public:
    Player(string);
    void setMoveChoice(string move);
    string getMoveChoice();
    ~Player();
};

#endif /* ROCK_PAPER_SCISSORSS_H */

Answer 1

您的类不需要析构函数，因为它从不显式分配任何资源。删除:

  Computer :: ~Computer(){
      delete this;
  }

和所有类似的功能。

请注意，即使您的类确实需要析构函数，析构函数也不应调用:

delete this;

因为这将由编译器发出的代码有效地为您执行。析构函数应该只删除您明确分配的资源，在被销毁的对象中，使用new。