嵌套 if 检查 C++

Question

也许我的代码很草率，但出于某种原因，它只是通过了所有嵌套的 If 检查，而没有停下来检查第一个以外的答案。此外，无论我给它一个 False 答案如何，第一个 If 检查都会继续检查嵌套的 If for True。我是 C++ 的新手，所以我缺少什么吗？

#include <iostream>
#include <string>

using namespace std;

int main()
{
    cout<<" \n";
    cout<<"UUU   UUU   RRRRRRR   TTTTTTTTT  HHH  HHH\n";
    cout<<"UUU   UUU   RRR  RRR     TTT     HHH  HHH\n";
    cout<<"UUU   UUU   RRRRRRR      TTT     HHHHHHHH\n";
    cout<<"UUU   UUU   RRR  RRR     TTT     HHH  HHH\n";
    cout<<" UUUUUUU    RRR   RRR    TTT     HHH  HHH\n";
    cout<<" \n";
    cout<<"           Created by: Illyduss\n";
    cout<<"                     2015\n";
    cout<<" \n";
    cout<<" \n";
    cout<<" \n";
    int account;
    cout<<"Do you already have an account? \n";
    cout<<"Type 'New' to create an account or enter your account name.\n";
    cout<<"Account Name: ";
    cin>> account;
    cin.ignore();
    if (account = "New" or "NEW" or "new"){
        string surname;
        cout<<" \n";
        cout<<"Account names serve as the Surname or Last name for\n";
        cout<<"all characters linked to said account. This is beca\n";
        cout<<"use of our unique gene pool system. Which will be c\n";
        cout<<"overed more in depth later on, but for now just thi\n";
        cout<<"nk of it like this, an account is a family tree for\n";
        cout<<"all of your characters.\n";
        cout<<" \n";
        cout<<"Please enter your desired Surname Name: ";
        cin>> surname;
        cin.ignore();
        if (surname.length() > 2){
            cout<<" \n";
            cout<<"You have chosen, '" << surname << "' as your surname, correct? ";
        }
        else {
            cout<<"That is too short, please choose another surname: ";
        }
    }
    else {
        cout<< "Welcome back, '" << account << "'!\n";
        cout<<"Please enter your password: ";
    }
    cin.get();
}

Answer 1

首先，您尝试使用 int 类型的对象来输入字符串，然后您尝试使用不正确的条件和赋值运算符 [=16] 将该对象与字符串文字进行比较=]

if (account = "New" or "NEW" or "new"){
            ^^

您应该将 account 定义为 std::string 类型，在这种情况下，条件可能类似于

if (account == "New" or account == "NEW" or account == "new"){

但无论如何最好将使用某些中间对象的 account 转换为大写。在这种情况下，条件看起来会更简单。

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