结构中的 C++ 函数和通过引用调用值

Question

所有-我已经对此进行了很多研究。我的程序编译没有错误，但结构中函数的值没有传递给程序。你能帮我弄清楚为什么他们不是吗？我包含了显示相关组件的代码的 sn-ps。主要是，我的代码如：“&allData::ConvertToC”没有从结构“allData”中的函数返回任何值。无论“allData.temperature”的输入如何，它都只会返回值 1。我知道该程序的所有组件，除了提到的那些都在工作。

代码 sn-ps：

//defining the struct

struct allData {
    char selection; 
    double centigrade; 
    double fahrenheit; 
    double temperature; 
    double ConvertToC (const double& temperature);
    double ConvertToF (const double& temperature);
} allData;

//adding data to the struct for the functions within the struct to use

cout << "Enter C for converting your temperature to Celsius, or enter F for converting your temperature to Fahrenheit, and press ENTER." << endl << endl;

cin >> allData.selection; 

cout << "Enter your starting temperature to two decimal places, and press ENTER." << endl << endl; 

cin >> allData.temperature; 

switch (allData.selection) {

//my attempt to reference the functions within the struct and the data in the struct, but it is not working and always returns a value of 1. 

case 'c': { &allData::ConvertToC; 

    cout << "Your temperature converted to Celsius is: " << &allData::ConvertToC   
    << endl << endl; 
    break; 
    }

case 'C': { &allData::ConvertToC; 

    cout << "Your temperature converted to Celsius is: " << &allData::ConvertToC    
    << endl << endl; 
    }
}


//Function definitions that are located in the struct. Do I define the functions in the normal way, like this, if they are located in the struct?

double allData::ConvertToF (const double& temperature) {

    double fahrenheit = 0;
    fahrenheit = temperature * 9 / 5 + 32;
    return fahrenheit; 

}


double allData::ConvertToC (const double& temperature) {

    double centigrade = 0;
    centigrade = (temperature - 32) * 5 /9; 
    return centigrade; 

}

Answer 1

您没有执行函数调用，您只是将函数指针传递给 cout 流。

我认为你真正想要的是这样的：

 cout << "Your temperature converted to Celsius is: " << allData.ConvertToC(allData.temperature) << endl; 

此外，您不需要在“ConvertToC”方法中通过引用传递，因为您并没有真正保存任何东西（“双精度”是 8 个字节宽，而引用/指针在 32 位系统上是 4 个字节，或 64 位系统上的 8 个字节）。

Answer 2

// Name of struct made distinct from its instance, for clarity.
struct AllData {
    char selection;
    double centigrade;
    double fahrenheit;
    double temperature;
    double ConvertToC ();
    double ConvertToF ();
} allData;

...

allData.selection = 'C';
allData.temperature = 74.5;

switch (allData.selection)
{
case 'c':
case 'C':
    cout << "Your temperature converted to Celsius is: " << allData.ConvertToC() << endl << endl;
    break;
}

...

double AllData::ConvertToF ()
{
    //double fahrenheit = 0;    Why not store the result in the struct?
    fahrenheit = temperature * 9 / 5 + 32;
    return fahrenheit;
}

double AllData::ConvertToC ()
{
    //double centigrade = 0;
    centigrade = (temperature - 32) * 5 / 9;
    return centigrade;
}