c ++编码选择开关代码的每个案例

Question

因此,即时通讯有一个项目,这是我的代码,但我在此代码,而不是只选择一个它选择每一种可能的情况下,这样帮助我有一个问题,这是我的代码也是不知道这是怎么发生,我还在寻找为什么会这样,但也许我需要专业的帮助,所以这是代码

#include <iostream>
#include <cmath>
#include <string>
using namespace std;

int main()
{
    char one, two, three, four;
    double a, b, c, d;



    cout << "What do you want to find on the square?" << endl;
    cout << "A. Area" << endl;
    cout << "B. Side" << endl;
    cout << "C. Diagonal" << endl;
    cout << "D. Perimeter" << endl;
    cin >> one ;
    one = toupper(one);
    switch (one)
        {
        case 'A':
            {   cout << "What is Given" << endl;
                cout << "S. Side" << endl;
                cout << "D. Diagonal" << endl;
                cout << "P. Perimeter" << endl;
                cin >> two;
                two = toupper(two);
                switch (two)
                    {   
                    case 'S':
                        {   
                            cout << "Enter Measure of the side." << endl;
                            cin >> a;
                            a= a*a;
                            cout << "The Answer is " << a << endl;

                        }
                    case 'D':
                        {
                            cout << "Enter Measure of the diagonal" << endl;
                            cin >> a;
                            a= pow( a/sqrt(2), 2);
                            cout << "The Answer is " << a << endl;

                        }
                    case 'P':
                        {
                            cout << "Enter measure of Perimeter" << endl;
                            cin >> c;
                            c= pow(c/4, 2);
                            cout << "The Answer is " << c << endl;
                        }
                    default :
                        {
                        }
                    }
            }
        }


    return 0;
}

Answer 1

你需要

 break

在每个案件之后.像这样:

 case 'S':
      {   
          cout << "Enter Measure of the side." << endl;
          cin >> a;
          a= a*a;
          cout << "The Answer is " << a << endl;
          break;
      }

对于每一个案例.否则它会经历所有情况而不会破坏,这就是你的情况