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c++ - 开关总是在猜数字游戏时输出默认值

转载 作者:行者123 更新时间:2023-11-27 23:21:03 25 4
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我正在尝试让一个程序运行,该程序会猜测用户在想哪个数字,但目前它每次都只输出默认的开关 Action 。请告诉我我做错了什么?谢谢!

srand(time(0));
int lowernum = 1;
int highernum = 1000;
int number=rand()%highernum + lowernum;
string letter = "";
int letternum = 0;

cout<<"\nOkay, think of a number between one and 1000
and I will try to guess it!\n";
cout<<"\nIs your number higher (h), lower (l)
or exactly (e): " << number << "\n";
cin>>letter;

if (letter == "h")
{
letternum = 1;
}
else if (letter == "l")
{
letternum = 2;
}
else if (letter == "e")
{
letternum = 3;
}

switch(letternum){
case'1': highernum = number; cout<<"\nIs your number higher (h),
lower (l) or exactly (e): " << number << "\n"; cin>>letter;
break;
case'2': lowernum = number; cout<<"\nIs your number higher (h),
lower (l) or exactly (e): " << number << "\n"; cin>>letter;
break;
case'3': cout<<"\nWahoooooo! I win! :D\n";
break;
default:cout<<"\nI don't understand what you just typed in.\n";
break;

}

最佳答案

letternum 是一个整数,您的 switch 语句使用字符(例如“1”),只需从您的 case 表达式中的字符中删除引号:

switch(letternum){
case 1:

等...

关于c++ - 开关总是在猜数字游戏时输出默认值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13020805/

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