Question

当我输入“ 1”（不带引号）作为输入时，它将带我到错误菜单（无效输入）。

重新措辞：当我运行该程序时，在初始形状提示菜单或“无效输入”菜单上，如果输入“ 1”，则无法继续选择。

有人对这里发生的事情有了解吗？

#include <iostream>
#include <string>
#include <limits>

using std::cout;
using std::cin;
using std::string;
using std::endl;

int main()
{
    string username;
    cout<<"Hello.\nMy name is Pythagoras.\nI will be helping you build shapes today.\n\nWhat is your name?\n"<<endl;
    cin>>username;
    int shapeselect;
    cout<<"Hello, "<<username<<".\nWhat shape would you like to build today?\n1)Rectangle/Square\n2)Triangle\n3)Random!\n(Please select a number.)\n"<<endl;
    cin>>shapeselect;

    while (shapeselect!=1||shapeselect!=2||shapeselect!=3)
    {
        cin.clear();
        std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
        cout<<"Sorry, "<<username<<", your entry was invalid.  Please select a valid number.\n";
        cout<<"What shape would you like to build today, "<<username<<"?\n1)Rectangle/Square!\n2)Triangle!\n3)Random!"<<endl;
        cin>>shapeselect;
    }

    if (shapeselect==1)
    {
        int width;
        cout<<"\nPlease enter the desired WIDTH of the rectangle, between 4 and 10.\n(Please select a number.)\n";
        cin>>width;
        cout<<"\nYou have selected a width of "<<width<<"."<<endl;
        int length;
        cout<<"\nPlease enter the desired LENGTH of the rectangle, between 4 and 10.\n(Please select a number.)\n";
        cin>>length;
        cout<<"\nYou have selected a length of "<<length<<"."<<endl;
        if (width==length)
        {
            cout<<"Please note that you have selected a width of "<<width<<" and a length of "<<width<<".\nNote that this shape will be a square.\n"<<endl;
        }
    }
    return 0;
}

Answer 1

据我了解，当shapeselect为1,2,3以外的某个数字时，您需要执行while循环。

要解决该问题，请在条件为while时用AND门替换所有OR门。

原因：因为，或门的任何一个输入为真时，输出为真。因此，如果shapeselect等于1，则它将不等于2和3，这使得shapeshelect!=2和shapeselect!=3的计算结果为true。无论您输入哪个数字，while循环都将始终运行。

菜单选择器不接受有效输入

1 个答案: