【问题标题】:How to store user string input in array, C++如何将用户字符串输入存储在数组中,C++
【发布时间】:2015-04-27 01:03:53
【问题描述】:

我正在构建一个选择你自己的冒险。

所以基本上,在用户输入他们的名字后,我希望 playerList switch 语句存储输入到 basePlayer[] 中的任何内容。

此外,如果您要运行此程序,则必须在输入姓名后单击向下箭头才能进入 playerList 菜单,因为它有点错误。

我们将不胜感激。

谢谢你, 迈克

#include <iostream>
#include <string>
#include <windows.h>
#include <cmath>
#include <cstring>
#include <cstdlib>

using namespace std;

int main(){
Beep(251.63,100);
Beep(329.63,100);
Beep(392,100);
Beep(251.63,100);
Beep(329.63,100);
Beep(392,100);
Beep(251.63,100);
Beep(329.63,100);
Beep(392,100);

string playerList[6] = {"Abbot", "Sear", "Hellion", "Vagabond", "Knave","##QUIT##"};
string cityList[4] = {"city1","city2","city3","city4"};
string spiritList[4] = {"spirit1","spirit2","spirit3","spirit4",};
string yesNo[2] = {"yes","no"};
string name;
string player;
string city;
string spirit;
string basePlayer[4];
    basePlayer[0] = name;
    basePlayer[1] = player;
    basePlayer[2] = city;
    basePlayer[3] = spirit;


int pointer = 0;

while(true){

cout << "What is your name?" << endl;
cin >> name;
cout << "your name is " << name << "?" << endl;
basePlayer[0] = name;
cout << basePlayer[0];


int pointer = 0;

while(true){
    system("cls");

    SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE), 5);
    cout << basePlayer[0]<<" please choose your Adventurer:\n*tab* for  description\n\n";

    for (int row = 0; row < 6; ++row){
        if(row == pointer){
            SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE), 11);
            cout << playerList[row] << endl;}

        else{
            SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE), 5);
            cout << playerList[row] << endl;}}

 while(true){

  if (GetAsyncKeyState(VK_UP) != 0){
        Beep(800,50);
    pointer -= 1;
    if (pointer == -1){
        pointer = 0;}
            break;}

 else if (GetAsyncKeyState(VK_DOWN) != 0){
        Beep(800,50);
    pointer += 1;
    if (pointer == 6){
        pointer = 0;}
            break;}

 else if (GetAsyncKeyState(VK_TAB) != 0){
        Beep(1200,50);
        Beep(1000,50);

    switch(pointer){
    case 0:{
        system("cls");
        cout << "AbbotFacts.\n\n*Enter* to become the Abbot\n*ArrowDown* to      return to TitleScreen";
        Sleep(1000);
    }
        break;
    case 1:{
        system("cls");
        cout << "SearFacts.\n\n*Enter* to become the Sear\n*ArrowDown* to  return to TitleScreen";
        Sleep(1000);
    }
        break;
    case 2:{
        system("cls");
        cout << "HellionFacts.\n\n*Enter* to become the Hellion\n*ArrowDown* to return to TitleScreen";
        Sleep(1000);
    }
        break;
    case 3:{
        system("cls");
        cout << "VagabondFacts.\n\n*Enter* to become the Vagabond\n*ArrowDown* to return to TitleScreen";
        Sleep(1000);
    }
        break;
    case 4:{
        system("cls");
        cout << "KnaveFacts.\n\n*Enter* to become the Knave\n*ArrowDown* to return to TitleScreen";
        Sleep(1000);
    }
        break;
    case 5:{return 0;}
        break;
        break;}}

else if (GetAsyncKeyState(VK_RETURN) != 0){
        Beep(1000,50);
        Beep(1200,50);


    switch(pointer){
    case 0:{
        system("cls");
        cout << "You have chosen the Abbot"<< endl;
        Sleep(1000);
    }

        break;
    case 1:{
        system("cls");
        cout << "You have chosen the Sear"<< endl;
        Sleep(1000);
    }
        break;
    case 2:{
        system("cls");
        cout << "You have chosen the Hellion"<< endl;
        Sleep(1000);
    }
        break;
    case 3:{
        system("cls");
        cout << "You have chosen the Vagabond" << endl;
        Sleep(1000);
    }
        break;
    case 4:{
        system("cls");
        cout << "You have chosen the knave"<< endl;
        Sleep(1000);
    }
        break;
    case 5:{return 0;
    break;}
    break;}}
 }
  Sleep(150);
 }
 }
 }

【问题讨论】:

  • 成功了吗?好的,我将添加它作为答案。请接受它。 :)
  • 好的,全部编辑完毕。
  • ^^ 顺便说一句,这可能会让我更容易看到我在说什么。
  • 如果您减少代码的大小以突出显示您遇到问题的代码部分,那将是非常有意义的。
  • 我建议您先用调试器开始解决该代码,如果有具体错误,请修改问题。

标签: c++ arrays variables input store


【解决方案1】:

尝试从您的程序中删除语句cin.get(),因为cin &gt;&gt; name 输入name,因此cin.get() 没有用处。

【讨论】:

  • 我回答完后还能再问你一个问题吗?
  • 当然。你想问什么?
  • 好的,那你到底有什么问题?
  • 我在“你的答案”中发布什么?我只是复制你的帖子吗?
  • 你不需要发布你的答案,因为你已经从这篇文章中得到了答案,我猜。您只需要在我的回答中单击赞成-反对票计数器下方的刻度线。
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