C++ 如何在不需要输入的情况下接收按键输入？ i=0； 而（i

您只需要存储输入并将其与所需案例进行比较。仔细考虑一下

i=0;
while(i<=20)
{
    nullchoice:
    system("CLS");
    //Advanced Information
    cout << "Your Statistics: " << endl;
    cout << endl << endl;
    cout << "   1  Strength: " << str << endl;
    cout << endl;
    cout << "   2  Vitality: " << vit << endl;
    cout << endl;
    cout << "   3  Agility: " << agi << endl;
    cout << endl;
    cout << "   4  Thaumaturgy: " << tha << endl;
    cout << endl << endl;
    cout << "Points Remaining: " << lvlskills << endl;
    cout << endl;
    cout << "Enter the number of the skill you wish to increase: ";

    //Applying points to skill attributes

    if(i<=19)
    {
          cin >> input;

          if(input==1)
                str+=1;
          else if(input==2)
               vit+=1;
          else if(input==3)
               agi+=1;
          else if(input==4)
               tha+=1;
          else
               goto nullchoice;

         lvlskills-=1; 
    }    
    i++;
    cout << endl << endl;                     
}

do
{
系统（“CLS”）；
//高级信息
一个相当可移植的解决方案是使用Curses库。你想用它做什么操作系统？这是针对Microsoft Windows的不，这不是一个重复的线程。其他解决方案包括如何接收按钮输入（getch、cin.get、pause>>nul等）。因此，如果我按“S”，它将继续，如果我按“R”我需要的是一个条件语句，这样程序就可以区分“R”和“S”其他问题中的答案解决了您的键盘输入问题。您还必须使用自己的大脑以您需要的方式使用收集的输入。这仍然需要用户按enter键。不..它没有..它经过测试..工作起来很有魅力。我现在明白了它的工作原理。谢谢！
do
{
    system("CLS");
    //Advanced Information
    cout << "Your Statistics: " << endl;
    cout << endl << endl;
    cout << "   1  Strength: " << str << endl;
    cout << endl;
    cout << "   2  Vitality: " << vit << endl;
    cout << endl;
    cout << "   3  Agility: " << agi << endl;
    cout << endl;
    cout << "   4  Thaumaturgy: " << tha << endl;
    cout << endl << endl;
    cout << "Points Remaining: " << lvlskills << endl;
    cout << endl;
    cout << "Enter the number of the skill you wish to increase: ";

    //Applying points to skill attributes
          char input;
          switch(input=getch()){
              case '1':
                str+=1;
                lvlskills-=1;
                break;
          case '2':
               vit+=1;
               lvlskills-=1;
               break;
          case '3':
               agi+=1;
               lvlskills-=1;
               break;
          case '4':
               tha+=1;
               lvlskills-=1;
         } 
    cout << endl << endl;                     
} while(lvlskills>0);