所以我做了一个DOS程序但是我的游戏总是在第二次运行到cin函数时崩溃。
#include <iostream>
#include <string>
#include <ctime>
#include <cstdlib>
using namespace std;
//call functions
int create_enemyHP (int a);
int create_enemyAtk (int a);
int find_Enemy(int a);
int create_enemyDef (int a);
// user information
int userHP = 100;
int userAtk = 10;
int userDef = 5;
string userName;
//enemy Information
int enemyHP;
int enemyAtk;
int enemyDef;
string enemies[] = {"Raider", "Bandit", "Mugger"};
int sizeOfEnemies = sizeof(enemies) / sizeof(int);
string currentEnemy;
int chooseEnemy;
// ACTIONS
int journey;
int test;
int main()
{
// main menu
cout << "welcome brave knight, what is your name? " ;
cin >> userName;
cout << "welcome " << userName << " to Darland" << endl;
//TRAVELING
MENU:
cout << "where would you like to travel? " << endl;
cout << endl << " 1.> Theives Pass " << endl;
cout << " 2.> Humble Town " << endl;
cout << " 3.> Mission HQ " << endl;
cin >> journey;
if (journey == 1)
{
// action variable;
string c_action;
cout << "beware your journey grows dangerous " << endl;
//begins battle
// Creating the enemy, HP ATK DEF AND TYPE. ;
srand(time(0));
enemyHP = create_enemyHP(userHP);
enemyAtk = create_enemyAtk(userAtk);
enemyDef = create_enemyDef(userDef);
chooseEnemy = find_Enemy(sizeOfEnemies);
currentEnemy = enemies[chooseEnemy];
cout << " Here comes a " << currentEnemy << endl;
cout << "stats: " << endl;
cout << "HP :" << enemyHP << endl;
cout << "Attack : " << enemyAtk << endl;
cout << "Defense : " << enemyDef << endl;
ACTIONS:
cout << "Attack <A> | Defend <D> | Items <I>";
cin >> c_action;
//if ATTACK/DEFEND/ITEMS choice
if (c_action == "A" || c_action == "a"){
enemyHP = enemyHP - userAtk;
cout << " you attack the enemy reducing his health to " << enemyHP << endl;
userHP = userHP - enemyAtk;
cout << "however he lashes back causing you to have " << userHP << "health left " << endl;
//end of ATTACK ACTION
}
最后一行“cin&gt;&gt; c_action崩溃。我使用另外两个页面。他们只是创建函数。这是一个编译器问题。为什么我的编译器在运行app后总是关闭。有没有办法停止它?
答案 0 :(得分:1)
一些提示: 我从不使用函数的前向声明(例如“int create_enemyHP(int a);”),如果我可以避免它们。如果这样做,那么代码中有两个位置必须正确才能使程序正常工作。如果总有“single source of truth”
,它会让生活更轻松您是否通过调试器运行此代码?它可以帮助您更快地找到问题。
答案 1 :(得分:0)
如果您的c_action
变量仅用作字符,我建议您使用char
变量,而不是string
。
您可能想尝试这种方式,如果您仍然遇到错误,可以给出
scanf("%c", &c_action); //assuming you used a char.
答案 2 :(得分:0)
我不明白程序是否在您输入“操作”之前或之后崩溃。因为如果它之前崩溃了,那么我认为你的问题是由输入缓冲区中的空格字符引起的。