Question

我正在使用在线编译器（运行Windows 8，并且无法让任何其他人工作）。我的代码运行并编译和完成，但它不起作用。假设询问用户风速和温度，将它们放入公式计算风寒，围绕它们（使用自制的圆形功能）并输出答案。但它什么都没给我。

感谢您的帮助。

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

int input (int wspeed, int tempature);
int calculate (int tempature, int wspeed, int windChill);
int trunkate (int windChill, double wchill);
int output (int tempature, int wspeed, int windChill);
int advisoryWarning (int windChill);


int main() 
{
int wspeed=0, tempature=0, windChill=0;
double wchill=0;
int input (int& wspeed, int& tempature);
int calculate (int& tempature, int& wspeed, int& windChill);
int trunkate (int& windChill, double& wchill);
int output (int& tempature, int& wspeed, int& windChill);
int advisoryWarning (int& windChill);
return 0;
}

int input (int& wspeed, int& tempature)
{
cout<<"This is the input function\nPlease input the tempature outside in fahrenheit."<<endl;
cin>>tempature;
cout<<"Please input wind speed in miles per hour."<<endl;
cin>>wspeed;

if(tempature<-40||tempature>40)
{
    cout<<"Please input a valid tempature between -40 and 40"<<endl;
    cin>>tempature;
}

if(wspeed<0||wspeed>60)
{
    cout<<"Please input a valid wind speed between 0 and 60 mph"<<endl;
    cin>>wspeed;
}
return 0;
}

int calculate (int& tempature, int& wspeed, double& wchill)
{
cout<<"\nThis is calculating\n";
wchill=35.74+0.6215*(tempature)-35.75*(pow(wspeed, 0.16))+0.4275*tempature*pow(wspeed, 0.16);
return wchill;
}

int trunkate (int& windChill, int& wchill)
{
//This is the round function
cout<<"\nThis is the trunkator function\nThis is the exact answer: " << wchill<<endl;
if(wchill<0)
    wchill=wchill-0.5;
else
    wchill=wchill+0.5;
cout<<"this is windchill plus or minus a half: "<<wchill<<endl;
windChill=wchill;
cout<<"this is the final answer: "<<windChill<<endl;
return windChill;
}

int output (int& tempature, int& wspeed, int& windChill)
{
cout<<"\nthis is the output function\n";
cout<<"Tempature entered: "<<tempature<<endl<<"Wind Speed entered: "<<wspeed<<endl<<"Wind Chill: "<<windChill<<endl<<endl;
cout<<"Calling the advisory output function...";
return 0;
}

int advisory (int& windChill)

{
if(windChill<-50)
    cout<<"Life threatening: Fostbite can occur within 5 minutes.";
else if (windChill<-30)
    cout<<"Danger: Frostbite can occur within 10 minutes.";
else if (windChill<-15)
    cout<<"Warning: Frostbite can occur within 30 minutes.";
else if (windChill<20)
    cout<<"Advisory: Frostbite can occur with extended exposure.";
else if (windChill<40)
    cout<<"Notice: Wear cold weather clothing as needed for comfort.";
else
    cout<<"No significant cold weather risk.";
return 0;
}

编辑：好的，所以我认为这是愚蠢的事情，对不起，我第一次参加编码课并遇到一些麻烦。所以拿出int＆amp;并用int替换它们，但现在我只输入运行。

我稍微更改了代码。

Answer 1

int main() 
{
    int wspeed=0, tempature=0, windChill=0;
    double wchill=0;
    input (int& wspeed, int& tempature);
    calculate (int& tempature, int& wspeed, int& windChill);
    trunkate (int& windChill, double& wchill);
    output (int& tempature, int& wspeed, int& windChill);
    advisoryWarning (int& windChill);
    return 0;
}

我为你修好了你的主要功能。当您调用其他函数时，不要调用函数的类型。当你创建一个名为

的函数时

int function(){}

int只是返回类型。当您调用另一个函数时，您只需要添加该函数的名称。在你的int main中，你只需要使用function（）;

进行调用

希望这会有所帮助。我不太擅长解释。

我无法从我的C ++程序中得到答案

1 个答案: