无法将双精度转换为字符串

Question

我一直在开发一个计算盒子体积的程序，使用 classes 帮助我理解如何使用 classes。作为程序的一部分，我想将对象的长度、宽度和高度转换为字符串以显示盒子的尺寸。当我运行我的主文件中的代码崩溃时。当我从 class 文件中运行时，我得到“无法将 Box::length 从双精度转换为 std::string。如何修复转换错误？

#include <iostream>
#include <stdexcept>
#include <sstream>
#include <iomanip>
#include <string>

using namespace std;

class Box
{
    public:
        double length;//length of the box
        double height;//height of the box
        double width;//with of the box

        Box(): length(1), height(1), width(1){}

        //Parameterized Constructor
        Box(double length, double width, double height);


        double getVolume(void);

        //Mutators
        void setLength(double leng);
        void setWidth(double wid);
        void setHeight(double hei);
        //Acessors
        string toString() const;
        string getLength(); 
        string getWidth();
        string getHeight();
};//end class   

//member function definitions

    double Box::getVolume(void)//get volume will cal and output the volume when called
    {
        return length * width * height;
    }
    void Box::setLength(double leng)
    {
        const double MIN_LENGTH = 0.1;//constants for min/max for range check and out_of_range exception
        const double MAX_LENGTH = 99;
        if (length > MAX_LENGTH || length < MIN_LENGTH)
        {
            stringstream strOut;//declare string stream

            strOut << "Length is out of range. Length must be between" << MIN_LENGTH << " and " << MAX_LENGTH << ".";//error msg
            throw out_of_range(strOut.str());
        }
        else
        {
            length = leng;// if length is within range, store it 
        }
    }
    string Box::getLength()
    {
        return length;
    }
    void Box::setWidth(double wid)
    {
        const double MIN_WIDTH = 0.1;//constants for min/max for range check and out_of_range exception
        const double MAX_WIDTH = 99;
        if (length > MAX_WIDTH || length < MIN_WIDTH)
        {
            stringstream strOut;//declare string stream

            strOut << "Width is out of range. Width must be between" << MIN_WIDTH << " and " << MAX_WIDTH << ".";//error msg
            throw out_of_range(strOut.str());
        }
        else
        {
            width = wid;// width is in range, store it
        }
    }   

    string Box::getWidth()
    {
        return width;
    }
    void Box::setHeight(double hei)
    {
        const double MIN_HEIGHT = 0.1;//constants for min/max for range check and out_of_range exception
        const double MAX_HEIGHT = 99;
        if (length > MAX_HEIGHT || length < MIN_HEIGHT)
        {
            stringstream strOut;//declare string stream

            strOut << "Height is out of range. Height must be between" << MIN_HEIGHT << " and " << MAX_HEIGHT << ".";//error msg
            throw out_of_range(strOut.str());
        }
        else
        {
            height = hei;// height is in range, store it
        }
    }   
    string Box::getHeight()
    {
        return height;
    }
    string Box::toString() const
    {
        stringstream strOut;
        strOut << "Length: " << getLength() << endl
        << "Width: " << getWidth() << endl <<
        "Height: " << getHeight() << endl;
        return strOut;
    }

Answer 1

您的 class 定义说 getLength() 将 return 一个字符串，但是您的 getLength() 函数实际上是 return 的宽度，它是一个双精度。您可能打算在 return 之前将长度转换为字符串。

您可以使用字符串库中的 to_string() 函数，并将您的 getLength() 更改为

return to_string(length)

Answer 2

此编译错误发生在 getWidth、getHeight、getLength 函数的 return 语句处。这是因为您将它们声明为 return 字符串，而不是 returned 宽度、高度和长度，它们是双精度的。编译器看到没有自动从double到string的转换。

要解决此问题，您只需将函数的 return 类型从字符串修复为双精度：

double getLength(); 
double getWidth();
double getHeight();

我注意到另一个错误：

doubletostring.cpp:104:25: error: reference to non-static member function must be called
strOut << "Length: " << getLength << endl

和 toString 方法中的类似错误。只需在末尾添加 () 即可将它们转换为函数调用：

strOut << "Length: " << getLength() << endl

无法将双精度转换为字符串

Could not convert double to string

c++

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