c++ - 如何在接受用户输入时运行相同的功能

Question

我想制作一个类似于游戏的基本控制台程序...所以问题是，当用户输入为 N/n 且没有任何错误时，我如何才能使相同的功能再次运行...这是我的代码。当我输入 N/n 时，它变成了 picture ...我正在使用 Visual Studio C++ 2015。提前谢谢你

#include "stdafx.h"
#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <string>
#include <iomanip>
using namespace std;

string name;
int age;
char prompt;

void biodata()
{
    cin.clear();
    cout << "Name : ";  getline(cin, name);
    cout << "Age : "; cin >> age;
}

void showBio()
{
    cin.clear();
    cout << "Thank you for providing your data...";
    cout << "\nPlease confirm your data...(Y/N)\n" << endl;

    //printing border
    cout << setfill('-') << setw(1) << "+" << setw(15) << "-" << setw(1) << "+" << setw(15) << "-" << setw(1) << "+" << endl;
    //printing student record
    cout << setfill(' ') << setw(1) << "|" << setw(15) << left << "Name" << setw(1) << "|" << setw(15) << left << "Age" << setw(1) << "|" << endl;
    //printing border
    cout << setfill('-') << setw(1) << "+" << setw(15) << "-" << setw(1) << "+" << setw(15) << "-" << setw(1) << "+" << endl;
    //printing student record
    cout << setfill(' ') << setw(1) << "|" << setw(15) << left << name << setw(1) << "|" << setw(15) << left << age << setw(1) << "|" << endl;
    //printing border
    cout << setfill('-') << setw(1) << "+" << setw(15) << "-" << setw(1) << "+" << setw(15) << "-" << setw(1) << "+" << endl;
    //printing student record

    cin >> prompt;

}

int main()
{
    cout << "Hi User, my name is Cheary. I'm your Computated Guidance (CG), nice to meet you..." << endl;
    cout << "Please provide your data..." << endl;
    biodata();
    showBio();

    if (prompt == 'Y' || 'y')
    {
        cout << "Thank you for giving cooperation...\nWe will now proceed to the academy..." << endl;
    }

    while (prompt == 'N' || 'n')
    {
        cout << "Please re-enter your biodata..." << endl;
        biodata();
        showBio();
    }

    system("pause");
    return 0;
}

Answer 1

不要使用全局变量!将它们作为参数或返回值传递。using namespace std; 是不好的做法。最好使用完整的限定名称。

你看到的问题是因为当你之前执行cin >> something;时，当涉及到getline(std::cin , 姓名);。然后您会立即得到一个空名称。

cin.clear() 并没有按照您的想法行事，参见documentation .

为了防止这种情况，您可以使用std::ws:

getline(cin >> std::ws, name);

这两个条件是错误的，而且总是正确的:

if (prompt == 'Y' || 'y') // equivalent to (prompt == 'Y' || 'y' not null) 
while (prompt == 'N' || 'n') // same with 'n' not null

你必须写两次prompt:

if (prompt == 'Y' || prompt == 'y')
while (prompt == 'N' || prompt == 'n')

使用 std::cin.ignore()而不是不可移植的 std::system("pause");。