在C++中将Char转换为Int的问题

Question

本网站上的第一个问题,希望我以可接受的方式发布代码.我正在努力使这个摇滚,纸张,剪刀游戏工作,但我很难通过"你已进入线",因为它在那之后终止.

我的教授说我的问题是当randNumGenerated是int类型时,userChoice是char类型.我尝试使用以下代码转换我的r,p和s值:char r ='a'; int a = r;

但是从编译器中得到了"重新定义:不同的基本类型"错误.我不知道该怎么做,因为将变量重新定义为int是我的意图.我觉得这一切都错了吗？任何帮助将不胜感激 !

int main()
{

    char userChoice;
    int computer;
    int randNumGenerated = 0;
    int r = 0;
    int p = 0;
    int s = 0;
    unsigned seed;



    cout << "Chose either rock, paper, or scissors" << endl;
    cout << "Let r,p,and s represent rock,paper,and scissors respectively " << endl;   

    cin >> userChoice;

    cout << "You entered: " << userChoice << endl;
    cout << " " << endl;

    seed = time(0);
    srand(seed);
    randNumGenerated = rand() % 3;

    r == 0;
    p == 1;
    s == 2;




    if ((userChoice == 0 && randNumGenerated == 2) || (userChoice == 1 && randNumGenerated == 0) || (userChoice == 2 && randNumGenerated == 1))
    {
        cout << "You win!";

    }

    if ((userChoice == 0 && randNumGenerated == 1) || (userChoice == 1 && randNumGenerated == 2) || (userChoice == 2 && randNumGenerated == 0))
    {
        cout << "You lose!";

    }

    if ((userChoice == 0 && randNumGenerated == 0) || (userChoice == 1 && randNumGenerated == 1) || (userChoice == 2 && randNumGenerated == 2))
    {
        cout << "It's a draw!";

    }






    return 0;

}

Answer 1

这三行没有效果(从字面上看,编译器应该警告你这个):

r == 0;
p == 1;
s == 2;

你可能正在寻找类似的东西:

int userNum = -1;
switch (userChoice) {
    case 'r':
        userNum = 0;
        break;
    case 'p':
        userNum = 1;
        break;
    case 's':
        userNum = 2;
        break;
}

但是,为什么你甚至根本不使用整数？如果您使用了字符表示,那么您的if语句将更容易阅读:

char randomChosen = "rps"[randNumGenerated];
if ((userChoice == 'r' && randomChosen == 's') || ...) {
     std::cout >> "You win!\n";
}

只要看一下这个,就可以看出它是Rock vs Scissor.在基于数字的代码中,我必须回顾转换表.