如何在其他类中使用参数化构造函数制作类的对象?

How to make object of a class with parameterized constructor in other class?

本文关键字：数制 函数 对象 参数 其他      更新时间：2023-10-16

我想在类中创建一个DataArea类的对象Area并在main函数中初始化数据。但是我的代码工作的唯一方法是初始化类中的数据Area。

另外，我不知道我是否正确制作了对象。请指导我。我的代码如下：

#include<iostream>
using namespace std;
class DataArea
{
public:
int radius, length, width, base, heigth;
DataArea(int l, int w, int b, int h, int r)
{
length = l;
width = w;
radius = r;
heigth = h;
base = b;
}
};
class Area
{
public:
DataArea* s = new DataArea(3, 4, 5, 6, 7);
float AreaCirle()
{
return 3.142 * s->radius * s->radius;
}
float AreaRectangle()
{
return s->length * s->width;
}
float AreaTraingle()
{
return (s->base * s->heigth) / 2;
}
};
class print_data : public Area
{
public:
void print()
{
cout << "Area of Circle is: " << AreaCirle() << endl;
cout << "Area of Rectangle is: " << AreaRectangle() << endl;
cout << "Area of Traingle is: " << AreaTraingle() << endl;
}
};
int main()
{
//DataArea da(3, 4, 5, 6, 7);
print_data m;
m.print();
}

如果你不在类之外使用它，你的DataArea基本上是绝对的Area。同样，print_data类也可以替换为operator<<重载。

以下是更新的代码，其中的注释将指导您完成。

#include <iostream>    
// DataArea (optionally) can be the part of Area  class
struct DataArea /* final */
{
float length, width, base, height, radius;
DataArea(float l, float w, float b, float h, float r)
: length{ l }  // use member initializer lists to initlize the members
, width{ w }
, base{ b }
, height{ h }
, radius{ r }
{}
};
class Area /* final */
{
DataArea mDataArea;  // DataArea  as member
public:
// provide a constructor which initialize the `DataArea` member
Area(float l, float w, float b, float h, float r)
: mDataArea{ l, w, b, h, r }  // member initializer 
{}
// camelCase naming for the functions and variables
// mark it as const as the function does not change the member
float areaCirle() const /* noexcept */
{
return 3.142f * mDataArea.radius * mDataArea.radius;
}
float areaRectangle() const /* noexcept */
{
return mDataArea.length * mDataArea.width;
}
float areaTraingle() const /* noexcept */
{
return (mDataArea.base * mDataArea.height) / 2.f;
}
// provide a operator<< for printing the results
friend std::ostream& operator<<(std::ostream& out, const Area& areaObject) /* noexcept */;
};
std::ostream& operator<<(std::ostream& out, const Area& areaObject) /* noexcept */
{
out << "Area of Circle is: " << areaObject.areaCirle() << "n";
out << "Area of Rectangle is: " << areaObject.areaRectangle() << "n";
out << "Area of Traingle is: " << areaObject.areaTraingle() << "n";
return out;
}
int main()
{
// now construct the Area object like this
Area obj{ 3, 4, 5, 6, 7 };
// simply print the result which uses the operator<< overload of the Area class
std::cout << obj;
}

输出：

Area of Circle is: 153.958
Area of Rectangle is: 12
Area of Traingle is: 15

在我看来，Area类对于你想要实现的目标来说是多余的。您可能应该将方法直接放在类DataArea。然后，您可以根据需要创建任意数量的DataArea对象...

喜欢这个：

class DataArea
{
public:
int radius, length, width, base, heigth;
DataArea(int l , int w , int b , int h , int r )
{
length = l;
width = w;
radius = r;
heigth = h;
base = b;
}

float AreaCirle()
{
return 3.142 * radius * radius;
}
float AreaRectangle()
{
return length * width ;
}
float AreaTraingle()
{
return (base * heigth)/2;
}
};
int main(int argc, char **argv)
{
DataArea area1 (1,2,3,4,5);
DataArea area2 (8,2,3,4,5);
std::cout << area1.AreaCirle() << std::endl;
std::cout << area2.AreaCirle() << std::endl;
}

您可能难以理解该概念的原因： 您正在定义一个类并实例化一个对象。有时这些术语可以互换使用，但在这种情况下，这是一个重要的区别。

如果您希望您的方法对其他类进行操作，则应创建接受该类作为参数的方法。否则，它是不必要的复杂。

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