小编The*_*Elf的帖子

所以我的朋友和我正在尝试制作一个基于文本的视频游戏,我一直在研究如何降低编程.这是我们迄今为止的c ++程序:

#include <iostream>
#include <stdio.h>

char Choice;

using namespace std;

int main()
{
    printf("You wake up to darkness.\n");
    printf("Confused and tired, you walk to the nearest house.\n");
    printf("You notice it's abandoned. What do you do?\n");
    printf("1. Walk Away.\n");
    printf("2. Jump.\n");
    printf("3. Open Door.\n");
    printf("Choose wisely.\n");
    cin >> Choice;

    if(Choice=1)
    {
        printf("The House seems to have a gravital pull on you. How strange.\n");
    }

    else if(Choice=2)
    {
        printf("Having Fun?\n");
    }

    return 0;
}

但是当我构建并运行它时,它将显示所有内容,但所有答案都将是if(Choice = 1)答案.我的程序中是否缺少某些需要或部分相互矛盾的东西？

 printf("What do you do?\n1. Walk Away.\n2. Jump.\n3. Open Door.\n\n");
 scanf("%d",&Choice);
 printf("\n\n\n");
 while(4<=Choice,Choice<=0);
 {
     printf("That is not a choice.\n");
     printf("What do you do?\n1. Walk Away.\n2. Jump.\n3. Open Door.\n\n");
     scanf("%d",&Choice);
     printf("\n\n\n");
 }

所以这是我的计划.它可以工作,但我想要做的是重复,直到输入1,2或3的答案.但无论答案是什么,它都会通过while循环然后继续,无论下一个选择.(另外,我确实宣布了"选择";我只是不想展示整个节目.)