Question

我要做的是为无效的char创建一个循环，例如：如果允许的字符是＆＃39; A＆＃39; a＆＃39;＆＃39; C＆＃39;＆＃39; c＆＃39;并且用户输入＆＃39; f＆＃39; 它会说它无效并要求他们输入一个新角色，直到他们输入一个有效的角色。

我也不确定它会进入以下代码的位置：

#include<iostream>
using namespace std;

int main (){
    int units;
    double priceA,priceB,priceC;
    char package, goAgain;
    priceA=15.95;
    priceB=23.95;
    priceC=45.95;

    do {
        cout << "Enter the Package chosen. Enter 'A' or 'a' for package A.n";
        cout << "Enter 'b' or 'B' for package B.n";
        cout << "Enter 'c' or 'C' for package C.n";
        cin >> package;
        cout << "Enter the number of message units";
        cin >> units;

        while(units>672){
            cout <<endl;
            cout <<"Invalid number,please enter a value from 0 to 672."<<endl;
            cout <<"Please Enter a new Number for units.";
            cin >>units;
        }

        switch (package){
            case 'A':
            case 'a':
            if(units>10){
                priceA=15.95+(units-10)*.20;
                priceB=23.95+(units-20)*.10;

                cout <<"The price of Package A is" << " $"<<priceA<< endl;
            }
            else if(units<=10){
                priceA=15.95;
                cout << "The price of Package A is" << " $"<<priceA<<endl;

            }
            if (priceA>priceB){
                cout <<"By switching to Package B you would save" << " $" << priceA-        priceB<<endl;
            }
            if (priceA>priceC){
                cout << "By switching to Package C you would save" << " $"<<priceA-             priceC<<endl;
            }
            break;
            case 'b':
            case 'B':
            if(units>10){
                priceA=15.95+(units-10)*.20;
                priceB=23.95+(units-20)*.10;

                cout <<"The price of Package B is" << " $"<<priceB<< endl;
            }
            else if(units<=10){
                priceB=23.95;
                cout << "The price of Package B is" << " $"<<priceB<<endl;
            }
            if (priceB>priceA){
                cout <<"By switching to Package A you would save" << " $" << priceB-    priceA<<endl;
            }
            if (priceB>priceC){
                cout << "By switching to Package C you would save" << " $"<<priceB-priceC<<endl;
            }
            break;
            case 'c':
            case 'C':
            if(units>10){
                priceA=15.95+(units-10)*.20;
                priceB=23.95+(units-20)*.10;

                cout <<"The price of Package C is" << " $"<<priceC<< endl;
            }
            else if(units<=10){
                priceC=45.95;
                cout << "The price of Package C is" << " $"<<priceC<<endl;
            }
            if (priceC>priceA){
                cout <<"By switching to Package A you would save" << " $" << priceC-priceA<<endl;
            }
            if (priceC>priceB){
                cout << "By switching to Package B you would save" << " $"<<priceC-priceB<<endl;
            }
            break;
        }

        cout<<"do you want to go again? y/n" << endl;
        cin >>goAgain;
    }
    while(goAgain=='y');
}

Answer 1

交换机可以有一个default分支，用于处理其他case分支中未处理的所有内容。您可以将代码放在那里，然后将整个switch放在do - while循环中，如下所示：

bool valid;
do {
    valid = true;

    switch (package) {
    case 'A':
        ⋮
    default:
        valid = false;
        cout << "Invalid bla bla " << package << ", enter it again" << endl;
        cin >> package;
        break;
    }
} while (!valid);

Answer 2

您可以尝试这样的事情：

bool valid = false;
while (!valid)
{
  const std::string valid_packages = "AaCc";
  char package;
  cin >> package;
  if (valid_packages.find(package) != std::string::npos)
  {
    valid = true;
  }
  else
  {
    cout << "\n"
         << "Invalid package name, try again.\n";
  }
}

有效的包裹字母放在一个字符串中。根据有效字母串检查用户的输入。如果找到用户的输入，则标志valid设置为true，结束循环。

为无效字符创建循环

2 个答案: