C++ Length类：unsigned int类型转换运算符及double构造器问题

作业要求 #

• 为Length类编写unsigned int类型转换运算符：将结果设置为len_inches对应十进制值的四舍五入结果
• 编写转换构造函数：实现从double类型创建Length对象

当前问题 #

已修复部分问题，但unsigned int类型转换运算符无法编译，只能使用普通int，不符合作业要求；同时转换构造函数存在逻辑错误。

相关代码 #

Length2.cpp #

#include "Length2.h"  

//*********************************************** 
// Operator double converts Length to a double  *
//***********************************************
Length::operator double() const
{
    return this->len_inches /12 + (this->len_inches %12) / 12.0;
}

//*********************************************** 
// Operator converts Length to a unsigned int  *
//***********************************************
Length::operator unsigned int() const
{
    return round(this->len_inches / 12 + (this->len_inches % 12) / 12.0);
}

//*********************************************
// Overloaded stream insertion operator <<    *
//*********************************************
ostream &operator<<(ostream& out, Length a)
{
    out << a.getFeet() << " feet, " << a.getInches() << " inches";
    return out;
}

Length2.h #

#pragma once 
#include <iostream>
using namespace std;

class Length
{ 
private:
    int len_inches;
public:
    Length(int feet, int inches)
    {
        setLength(feet, inches);
    }
    Length(int inches){ len_inches = inches; }
    Length(double feet){ len_inches = feet * 144; }
    
    int getFeet() const{ return len_inches / 12; }
    int getInches() const { return len_inches % 12; }
    void setLength(int feet, int inches)
    {
        len_inches = 12 *feet + inches;
    }
    // Type conversion operators
    operator double() const;
    operator int() const;
    
    // Overloaded stream output operator
    friend ostream &operator<<(ostream &out, Length a);

    //Convert Constructor from double
    Length(double dValue)
    {
        value = dValue;
    }
    double getValue() const { return value; }
};

main.cpp #

// This program demonstrates the type conversion operators for
// the Length class. 
#include "Length2.h"   

#include <iostream>
#include <string>
#include <fstream> //fstream, ifstream, ofstream
#include <cmath>
using namespace std;

int main()
{
   int ui = 0;
    Length distance(4, 6);
    ui = distance;
    cout << ui << " feet." << endl;

    Length distance1(4.5);
    double ui2 = -1;
    ui2 = distance1;
    cout << ui2 << " inches." << endl;


   return 0;
}

解决方案 #

1. 修复unsigned int转换运算符编译问题 #

• 在Length2.h的类型转换运算符声明中，添加unsigned int版本的声明：

// Type conversion operators
operator double() const;
operator int() const;
operator unsigned int() const; // 新增此行

• 转换运算符实现中，将round返回的double显式转换为unsigned int，同时简化计算逻辑：

Length::operator unsigned int() const
{
    double total_feet = this->len_inches / 12.0;
    return static_cast<unsigned int>(round(total_feet));
}

2. 修复double转换构造函数问题 #

• 类中不存在value成员变量，所有构造函数应操作核心成员len_inches
• 合并重复的double参数构造函数，明确逻辑为将英尺数转换为总英寸数：

// 替换原有两个double参数构造函数
Length(double feet)
{
    len_inches = static_cast<int>(round(feet * 12));
}

• 删除类中错误的value相关代码（包括double getValue() const和构造函数内的value = dValue;）

3. 调整main函数测试逻辑 #

若要测试unsigned int转换运算符，需将对应变量类型改为unsigned int：

unsigned int ui = 0;
Length distance(4, 6);
ui = distance;
cout << ui << " feet." << endl;

内容的提问来源于stack exchange，提问作者Anonymous