需要一个解决方案 C++ 循环和条件

Question

当有人从选项中输入超出范围的数字时，我希望程序显示错误......

#include "stdafx.h"
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <conio.h>
#include <process.h>
#define PI 3.14159265359
using namespace std;

int main()
{
     int a, b, c, d, e, f;
     long double loga, logarithm, sine, cosine,tan;
     char ch, ch1;
     cout << "\nMy Friend Calci:-";
     cout << "\n1. Addition (1st + 2nd)";
     cout << "\n2. Subtraction (1st - 2nd)";
     cout << "\n3. Multiplication (1st * 2nd)";
     cout << "\n4. Division (1st/2nd)";
     cout << "\n5. Logarithm";
     cout << "\n6. Natural Sine";
     cout << "\n7. Natural Cosine";
     cout << "\n8. Natural Tangent";
     cout << "\n9. Exit";
     do
     {
         cin >> ch;
         if (ch == '1' || ch == '2' || ch == '3' || ch == '4')
         {
          cout << "\nEnter the first and the second numbers respectively:-";
          cin >> a >> b;
         }
         else if (ch == '5' || ch == '6' || ch == '7' || ch == '8')
         {
          cout << "\nEnter the angle (in Radians) or the number you want to calculate Log/Sine/Cosine of:-";
           cin >> loga;
         }
        switch (ch)
        {
        case '1':c = a + b;
            cout << "Sum =" << c;
            break;
        case '2':d = a - b;
            cout << "Difference =" << d;
            break;
        case '3':e = a*b;
            cout << "Product =" << e;
            break;
        case '4':f = a / b;
            cout << "Quotient =" << f;
            break;
        case '5':logarithm = log(loga);
            cout << "log(%d) =" << logarithm;
            break;
        case '6':sine = sin(loga*PI / 180);
            cout << "sin(%d) =" << sine;
            break;
        default:cout << "Wrong Choice My Friend... Now DIE!!!!!";
            cout << "\nOr enter a valid one";
            break;
        }
    cout << "\nWant to Enter MORE (y/n) ?????";
    cin >> ch1;
    if (ch1 == 'y' || ch1 == 'Y')
        cout << "Then enter a choice:";
} while (ch1 == 'y' || ch1 == 'Y');
return 0;
}

当有人输入 1 到 9 的任何其他数值时，我需要程序显示错误消息，同时显示 1 到 4 和 5 到 8 的两条不同消息，如代码所示，如​​果可能的话然后至少没有。循环......（并不是说已经有更少了......呵呵......）

更新！！！！！！ 经过一些调整，如果我输入 156，现在发生的事情是 dat，它将 1 作为输入并要求输入，如果我输入 1，那么它会给我 56 和 1 的总和，依此类推....

Answer 1

do
     {
         cin >> ch;

添加：

         if (ch < '1' || ch > '9')
         {
             cout << "Unknown option " << ch << "! Please retry: " << endl;
             ch1 = 'y';
             continue;
         }

         if (ch == '1' || ch == '2' || ch == '3' || ch == '4')
         {
          cout << "\nEnter the first and the second numbers respectively:-";
          cin >> a >> b;
         }

或者如果你想保留当前格式，那么而不是

    default:cout << "Wrong Choice My Friend... Now DIE!!!!!";
        cout << "\nOr enter a valid one";
        break;

这样写：

default:cout << "Wrong Choice My Friend... Now DIE!!!!!";
    cout << "\nOr enter a valid one ";
    ch1 = 'y';
    continue;

Answer 2

只是一个快速的建议。您可以将输入类型从 char ch 更改为 int ch。 char 是一个字符。如果输入 10，则只取“1”。如果输入 22。它只需要 '2'。输入的其余部分用于该功能。例如，如果您输入 22。选择选项“2”。 a = 2。 b = 等待用户。 此外，您没有提供正确的退出功能。您可以添加到您的开关条件。

int a, b, c, d, e, f;
long double loga, logarithm, sine, cosine, tan;
char ch1;
int ch; // int ch instead of char ch
cout << "\nMy Friend Calci:-";
cout << "\n1. Addition (1st + 2nd)";
cout << "\n2. Subtraction (1st - 2nd)";
cout << "\n3. Multiplication (1st * 2nd)";
cout << "\n4. Division (1st/2nd)";
cout << "\n5. Logarithm";
cout << "\n6. Natural Sine";
cout << "\n7. Natural Cosine";
cout << "\n8. Natural Tangent";
cout << "\n9. Exit";
do
{
    cin >> ch;
    if (ch == 1 || ch == 2 || ch == 3 || ch == 4)
    {
        cout << "\nEnter the first and the second numbers respectively:-";
        cin >> a >> b;
    }
    else if (ch == 5 || ch == 6 ) // remove 7 and 8 because there is not matching function in switch statement. Pointless to accept user input.
    {
        cout << "\nEnter the angle (in Radians) or the number you want to calculate Log/Sine/Cosine of:-";
        cin >> loga;
    }
    switch (ch)
    {
    case 1:c = a + b; // change to int instead of char input for all cases
        cout << "Sum =" << c;
        break;
    case 2:d = a - b;
        cout << "Difference =" << d;
        break;
    case 3:e = a*b;
        cout << "Product =" << e;
        break;
    case 4:f = a / b;
        cout << "Quotient =" << f;
        break;
    case 5:logarithm = log(loga);
        cout << "log(%d) =" << logarithm;
        break;
    case 6:sine = sin(loga*PI / 180);
        cout << "sin(%d) =" << sine;
        break;
    case 9: exit(0); // exit your program 
        break;
    default:cout << "Wrong Choice My Friend... Now DIE!!!!!";
        cout << "\nOr enter a valid one";
        break;
    }
    cout << "\nWant to Enter MORE (y/n) ?????";
    cin >> ch1;
    if (ch1 == 'y' || ch1 == 'Y')
        cout << "Then enter a choice:";
} while (ch1 == 'y' || ch1 == 'Y');