该代码不会在其中显示任何数字输入.如何使它显示任何输入

the code will not display any number input into it. how to make it so it will display whatever is the input?

本文关键字：任何 显示 何使它 输入 数字输入 代码 在其中      更新时间：2023-10-16

>我正在使用天，小时和分钟在c ++中做一个3d数组，用户可以输入任何数字，它将显示为day：hour：minute问题是代码不会显示任何给定的数字。

我尝试将 for 循环中的天、小时和分钟设置为 1 并且有效，但仅当输入 1 时，任何其他数字都不会运行

int main()
{
    float temp[365][24][23] = { 0.0 };// this sets the array to have no more then 365 days-change temp to days
    char stop = 'n';
    int hour = 1;
    int min = 1;
    int dayz = 1; // will change i to temp
    do
    {
        std::cout << "please give me a day from 1-365" << std::endl;//tells users that 1-365 days are needed
        std::cin >> dayz;
        std::cout << "plese give me a number from 1-24 hours  " << std::endl;//tells users that 1-365 days are neede
        std::cin >> hour;
        std::cout << "please give me the minute" << std::endl;//tells users that 1-365 days are needed
        std::cin >> min;
        std::cout << "enter the temperature for today" << " = "; //prompt users to keep on giving temperatures
        std::cin >> temp[dayz][hour][min];
        std::cout << "If you want to end the program type N.If you dont then type any number" << std::endl; // tells users after date and temperatures are entered if the they want to stop type 0 and if not press anynumber
        std::cin >> stop;
    } while (stop != 'N' && stop != 'n');
    for (dayz = 0; dayz <=364; ++dayz)// makes the calendar and keeps it less then 366 issue is that it will show the same date multiple time
    {
        for (hour = 0; hour <=23; ++hour)// keeps the calendar orderly      
        {
            for (min = 0; min <= 22; ++min)
            {
                if (temp[dayz][hour][min] != 0.0)
                {
                    std::cout << dayz << ":" << hour << ":" << min << " = " << temp[dayz][hour][min] << "/t "; //will display all the input
                }
            }
            return 0;
        }
    }
}

你在内部循环之后写return 0;，我想这不是你想要的。你应该使用std::string stop = "n";并比较stop != "N" && stop != "n"

试试这个。

int main(){
    float temp[365][24][23] = { 0.0 };// this sets the array to have no more then 365 days-change temp to days
    char stop = 'n';
    int hour = 1;
    int min = 1;
    int dayz = 1; // will change i to temp
    do{
        std::cout << "please give me a day from 1-365" << std::endl;//tells users that 1-365 days are needed
        std::cin >> dayz;
        std::cout << "plese give me a number from 1-24 hours  " << std::endl;//tells users that 1-365 days are neede
        std::cin >> hour;
        std::cout << "please give me the minute" << std::endl;//tells users that 1-365 days are needed
        std::cin >> min;
        std::cout << "enter the temperature for today" << " = "; //prompt users to keep on giving temperatures
        std::cin >> temp[dayz][hour][min];
        std::cout << "If you want to end the program type N.If you dont then type any number" << std::endl; // tells users after date and temperatures are entered if the they want to stop type 0 and if not press anynumber
        std::cin >> stop;
    } while (stop != 'N' && stop != 'n');
    for (dayz = 0; dayz <=364; ++dayz)// makes the calendar and keeps it less then 366 issue is that it will show the same date multiple time
    {
        for (hour = 0; hour <=23; ++hour)// keeps the calendar orderly      
        {
            for (min = 0; min <= 22; ++min){
                if (temp[dayz][hour][min] != 0.0)
                    std::cout << dayz << ":" << hour << ":" << min << " = " << temp[dayz][hour][min] << "/t "; //will display all the input
            }
        }
    }
     return 0;
}

所以，我找到了一种方法来修复它，只需摆脱 for 循环并将 min 的 for 循环转换为 if 语句

。 例如。 
int main()
{
    float temp[364][23][59] = { 0.0 };//-1073741571 i get this error when i put minute in 59
    char stop = 'n';
    int hour = 1;
    int min = 1;
    int dayz = 1; // when i dont get the error code the program doesnt show
    do
    {
        std::cout << "please give me a day from 1-365" << std::endl;//tells users that 1-365 days are needed
        std::cin >> dayz;
        std::cout << "plese give me a number from 1-24 hours  " << std::endl;//tells users that 1-365 days are neede
        std::cin >> hour;
        std::cout << "please give me the minute" << std::endl;//tells users that 1-365 days are needed
        std::cin >> min;
        std::cout << "enter the temperature for today" << " = "; //prompt users to keep on giving temperatures
        std::cin >> temp[dayz][hour][min];
        std::cout << "If you want to end the program type N.If you dont then type any number" << std::endl; // tells users after date and temperatures are entered if the they want to stop type 0 and if not press anynumber
        std::cin >> stop;
    } while (stop != 'N' && stop != 'n');
        if  (min < 61 )
            {
                    std::cout << dayz << ":" << hour << ":" << min << " = " << temp[dayz][hour][min] << "/t "; //will display all the input
                    return 0;
            }
}
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