在平行数组中重新启动循环

Question

对于我的分配，我不能允许用户在数组中输入rainfall的负值。我想再次重新启动循环，好像他们没有输入任何要开始的东西，然后让他们重试。

尝试它时，我第一次输入一个负值，它将在一月份重新启动，但是在此之后，我可以输入更多的负面值，它只是要求我继续进入剩下的几个月。如果我继续给出负数，我希望它继续重新启动，直到开始进入正数为止。然后，那是我显示总和平均的时候。

{
    const int SIZE = 12;
    string months[SIZE] = { "January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December" };
    string Month[SIZE];
    double Rainfall[SIZE];
    int counter;
    double totalRainfall = 0.0;
    double averageRainfall = 0.0;
    for (counter = 0; counter < SIZE; counter++)
    {
        cout << "Enter the total rainfall for " << months[counter] << ": ";
        cin >> Rainfall[counter];
        if (Rainfall[counter] < 0.0)
        {
            cout << "Sorry, can't be a negative!" << endl;
            do
            {
                for (counter = 0; counter < SIZE; counter++)
                {
                    cout << "Enter the total rainfall for " << months[counter] << ": ";
                    cin >> Rainfall[counter];
                }
            } while (Rainfall[counter] < 0.0);
        }
        averageRainfall = totalRainfall / SIZE;
        cout << "Total rainfall for the whole year was: " << totalRainfall << setprecision(3) << endl;
        cout << "The average inches of rainfall was: " << averageRainfall << setprecision(3) << endl;
        system("pause");
        return 0;
    }
}

Answer 1

这是您要完成的工作的基本工作示例。与其添加然后检查，我先检查了值，然后将值添加到数组中。我创建了一个简单的功能，以确保输入是有效的数字。如果不是这样，只需将计数器重置回0即可从头开始。

编辑：添加尝试和捕获块以确保输入是正确的double

#include <iostream>
#include <string>
bool is_valid_num(std::string str){
    //check if its a number
    for(char c : str){
        if((c < 48 || c > 57 ) && c != 46) return false;
    }
    //if you want it to also not be equal to 0
    // if(std::stoi(str) < 0) return false;
    return true;
}
int main() {
    const int SIZE = 12;
    std::string months[] = { "January", "February", "March", "April", "May", "June", 
                        "July", "August", "September", "October", "November", "December" };

    std::string temp_input;
    double rainfall[12]; 
    double total_rainfall = 0.0;
    for(int i = 0; i < SIZE; i++){
        std::cout << "enter rainfall for " << months[i] << ": ";
        std::cin >> temp_input;
        //restart back to january
        if(!(is_valid_num(temp_input))) i = -1;
        rainfall[i] = std::stod(temp_input);
        total_rainfall += std::stod(temp_input);
    }
    std::cout << "avg rainfall: " << total_rainfall / 12.0 << "n";
    std::cout << "total rainfall " << total_rainfall << "n";
}

Answer 2

好吧，那是您陷入困境的时循环，当您第二次放置负值时，您已经在循环时，因此不会打印对不起....而不是从循环中重新启动。计数器= 0，宁愿这样做

if (Rainfall[counter] < 0.0)
    {
        cout << "Sorry, can't be a negative!" << endl;
        counter--;// either put zero if you want to restart whole loop else decrement
        continue;
    }