编写此代码的更有效方法是什么? P>我是C++的新手,我想学习继承。我写了一个程序,只需打印出不同形状的周长,即正方形和三角形。不幸的是,我这样做的结果是一些重复的代码。我的意思是: #include <iostream> class Shape { public: int sides; int sideLength; }; class Square : public Shape { public: int sides = 4; int sideLength = 6; void calculatePerimeter() { int perimeter = sideLength * sides; std::cout << "perimeter length: " << perimeter << std::endl; } }; class Triangle : public Shape { public: int sides = 3; int sideLength = 4; void calculatePerimeter() { int perimeter = sideLength * sides; std::cout << "perimeter length: " << perimeter << std::endl; } }; int main() { Square s; std::cout << "number of sides a square has: " << s.sides << std::endl; Triangle t; std::cout << "number of sides a triangle has: " << t.sides << std::endl; s.calculatePerimeter(); t.calculatePerimeter(); }
如您所见,calculatePerimeter函数正在正方形和三角形类中使用。将其放入Shape类中并调用它会导致输出为0。如何提高效率,使我不必为我创建的每个形状类复制/粘贴它?此代码: 内边=3; 内边长=4 正在使用局部范围视图创建新变量。在不运行代码的情况下,我可以想象,如果您简单地去掉两行前面的“int”,应该可以解决您的问题 编辑:当然,将它们放入构造函数中 编辑:好的,合并代码编写此代码的更有效方法是什么? P>我是C++的新手,我想学习继承。我写了一个程序,只需打印出不同形状的周长,即正方形和三角形。不幸的是,我这样做的结果是一些重复的代码。我的意思是: #include <iostream> class Shape { public: int sides; int sideLength; }; class Square : public Shape { public: int sides = 4; int sideLength = 6; void calculatePerimeter() { int perimeter = sideLength * sides; std::cout << "perimeter length: " << perimeter << std::endl; } }; class Triangle : public Shape { public: int sides = 3; int sideLength = 4; void calculatePerimeter() { int perimeter = sideLength * sides; std::cout << "perimeter length: " << perimeter << std::endl; } }; int main() { Square s; std::cout << "number of sides a square has: " << s.sides << std::endl; Triangle t; std::cout << "number of sides a triangle has: " << t.sides << std::endl; s.calculatePerimeter(); t.calculatePerimeter(); },c++,inheritance,C++,Inheritance,如您所见,calculatePerimeter函数正在正方形和三角形类中使用。将其放入Shape类中并调用它会导致输出为0。如何提高效率,使我不必为我创建的每个形状类复制/粘贴它?此代码: 内边=3; 内边长=4 正在使用局部范围视图创建新变量。在不运行代码的情况下,我可以想象,如果您简单地去掉两行前面的“int”,应该可以解决您的问题 编辑:当然,将它们放入构造函数中 编辑:好的,合并代码 class Triangle : public Shape { public: Trian
class Triangle : public Shape {
public:
Triangle ()
{
sides = 3;
sideLength = 4;
}
void calculatePerimeter() {
int perimeter = sideLength * sides;
std::cout << "perimeter length: " << perimeter << std::endl;
}
};
根据您试图理解的内容,其他答案是更好的问题解决方案。这很简单,只需将参数放入CalculatePerimeter方法,但这当然不会测试您对某个范围内变量的理解,这似乎是您的绊脚石。此代码:
内边=3;
内边长=4
正在使用局部范围视图创建新变量。在不运行代码的情况下,我可以想象,如果您简单地去掉两行前面的“int”,应该可以解决您的问题
编辑:当然,将它们放入构造函数中
编辑:好的,合并代码
class Triangle : public Shape {
public:
Triangle ()
{
sides = 3;
sideLength = 4;
}
void calculatePerimeter() {
int perimeter = sideLength * sides;
std::cout << "perimeter length: " << perimeter << std::endl;
}
};
根据您试图理解的内容,其他答案是更好的问题解决方案。这很简单,只需将参数放入CalculatePerimeter方法,但这当然不会测试您对某个范围内变量的理解,这似乎是您的绊脚石。当您使用继承时,您可以在基类中考虑常见行为和状态,并使派生类使用其特定功能扩展此状态和行为 从形状上拆下侧面和边长构件。相反,在类形状中提供返回这些值的虚拟函数,并让派生类重写它们 Move方法计算类形状的周长,并使用getSides和getSideLength实现它
#include <iostream>
class Shape {
public:
virtual int getSides() = 0;
virtual int getSideLength() = 0;
void calculatePerimeter() {
int perimeter = getSideLength() * getSides();
std::cout << "perimeter length: " << perimeter << std::endl;
}
};
class Square : public Shape {
public:
virtual int getSides() {
return 4;
}
virtual int getSideLength() {
return 6;
}
};
class Triangle : public Shape {
public:
virtual int getSides() {
return 3;
}
virtual int getSideLength() {
return 4;
}
};
这就是所谓的模板方法设计模式,它不与C++的模板函数概念混淆,它是与此语言相关的完全无关的概念。您可以在基类中考虑常见行为和状态,并使派生类使用其特定功能扩展此状态和行为
从形状上拆下侧面和边长构件。相反,在类形状中提供返回这些值的虚拟函数,并让派生类重写它们 Move方法计算类形状的周长,并使用getSides和getSideLength实现它#include <iostream>
class Shape {
public:
virtual int getSides() = 0;
virtual int getSideLength() = 0;
void calculatePerimeter() {
int perimeter = getSideLength() * getSides();
std::cout << "perimeter length: " << perimeter << std::endl;
}
};
class Square : public Shape {
public:
virtual int getSides() {
return 4;
}
virtual int getSideLength() {
return 6;
}
};
class Triangle : public Shape {
public:
virtual int getSides() {
return 3;
}
virtual int getSideLength() {
return 4;
}
};
这就是所谓的模板方法设计模式,它不与C++的模板函数概念混淆,它是与此语言无关的概念。
可以直接访问基类成员,也可以初始化它们。class Shape {
public:
int sides;
int sideLength;
Shape(int m,int n)
{
sides =m;
sideLength = n;
}
};
class Square : public Shape {
public:
Square(int j,int k);
int calculatePerimeter() {
int perimeter = sides * sideLength ;
std::cout << "perimeter length: " << perimeter << std::endl;
return perimeter;
}
};
Square::Square(int j,int k):Shape(j,k)
{
}
您可以访问直接的基类成员并初始化它们
class Shape {
public:
int sides;
int sideLength;
Shape(int m,int n)
{
sides =m;
sideLength = n;
}
};
class Square : public Shape {
public:
Square(int j,int k);
int calculatePerimeter() {
int perimeter = sides * sideLength ;
std::cout << "perimeter length: " << perimeter << std::endl;
return perimeter;
}
};
Square::Square(int j,int k):Shape(j,k)
{
}
您需要一个用于Shape和传递派生构造函数中的变量的构造函数。然后可以将calculatePerimiter移动到父类中
class Shape {
public:
Shape(int n_sides, int side_length) : sides(n_sides), sideLength(side_length) {}
void calculatePerimeter() {
int perimeter = sideLength * sides;
std::cout << "perimeter length: " << perimeter << std::endl;
}
int sides;
int sideLength;
};
class Square : public Shape {
public:
Square() : Shape(4,6) {}
};
class Triangle : public Shape {
public:
Triangle() : Shape(3, 4) {}
};
您需要一个用于Shape和传递派生构造函数中的变量的构造函数。然后可以将calculatePerimiter移动到父类中
class Shape {
public:
Shape(int n_sides, int side_length) : sides(n_sides), sideLength(side_length) {}
void calculatePerimeter() {
int perimeter = sideLength * sides;
std::cout << "perimeter length: " << perimeter << std::endl;
}
int sides;
int sideLength;
};
class Square : public Shape {
public:
Square() : Shape(4,6) {}
};
class Triangle : public Shape {
public:
Triangle() : Shape(3, 4) {}
};
这应该能回答你的问题。另外,我建议将sides和sideLength的值作为参数传递给函数调用,以实现一般用法
#include <iostream>
class Shape {
protected void calculatePerimeter(int sides, int sideLength) {
int perimeter = sideLength * sides;
std::cout << "perimeter length: " << perimeter << std::endl;
}
};
class Square : public Shape {
int sides = 4;
int sideLength = 6;
public void calculatePerimeter()
{
Shape::calculatePerimeter(sides, sideLength);
}
};
class Triangle : public Shape {
int sides = 3;
int sideLength = 4;
public void calculatePerimeter()
{
Shape::calculatePerimeter(sides, sideLength);
}
};
int main() {
Square s;
std::cout << "number of sides a square has: " << s.sides << std::endl;
Triangle t;
std::cout << "number of sides a triangle has: " << t.sides << std::endl;
s.calculatePerimeter();
t.calculatePerimeter();
}
这应该能回答你的问题。另外,我建议将sides和sideLength的值作为参数传递给函数调用,以实现一般用法
#include <iostream>
class Shape {
protected void calculatePerimeter(int sides, int sideLength) {
int perimeter = sideLength * sides;
std::cout << "perimeter length: " << perimeter << std::endl;
}
};
class Square : public Shape {
int sides = 4;
int sideLength = 6;
public void calculatePerimeter()
{
Shape::calculatePerimeter(sides, sideLength);
}
};
class Triangle : public Shape {
int sides = 3;
int sideLength = 4;
public void calculatePerimeter()
{
Shape::calculatePerimeter(sides, sideLength);
}
};
int main() {
Square s;
std::cout << "number of sides a square has: " << s.sides << std::endl;
Triangle t;
std::cout << "number of sides a triangle has: " << t.sides << std::endl;
s.calculatePerimeter();
t.calculatePerimeter();
}
这将是我的方法为编译器,支持C++11
#include <iostream>
#include <string>
#include <vector>
using namespace std;
template<class Seq>
void purge(Seq& s) {
for (auto x: s) {
delete x;
x = nullptr;
}
}
class Shape {
public:
virtual void calculatePerimeter() = 0; // to avoid object slicing
virtual ~Shape() {};
};
class Square : public Shape {
int m_sides{0};
int m_sideLength{0};
public:
Square() = delete; // suppress default constructor
Square(int sides, int sideLength): m_sides(sides), m_sideLength(sideLength) {}
~Square() {}
void calculatePerimeter() {
int perimeter = m_sideLength * m_sides;
cout << "Square perimeter length: " << perimeter << endl;
}
};
class Triangle : public Shape {
int m_sides{0};
int m_sideLength{0};
public:
Triangle() = delete; // suppress default constructor
Triangle(int sides, int sideLength): m_sides(sides), m_sideLength(sideLength) {}
~Triangle() {}
void calculatePerimeter() {
int perimeter = m_sideLength * m_sides;
cout << "Triangle perimeter length: " << perimeter << endl;
}
};
int main() {
vector<Shape*> vec;
vec.push_back(new Square(4, 7));
vec.push_back(new Triangle(3, 13));
for (auto x: vec) {
x->calculatePerimeter();
}
// clean up
purge(vec);
system("pause");
return 0;
}
这将是我的方法为编译器,支持C++11
#include <iostream>
#include <string>
#include <vector>
using namespace std;
template<class Seq>
void purge(Seq& s) {
for (auto x: s) {
delete x;
x = nullptr;
}
}
class Shape {
public:
virtual void calculatePerimeter() = 0; // to avoid object slicing
virtual ~Shape() {};
};
class Square : public Shape {
int m_sides{0};
int m_sideLength{0};
public:
Square() = delete; // suppress default constructor
Square(int sides, int sideLength): m_sides(sides), m_sideLength(sideLength) {}
~Square() {}
void calculatePerimeter() {
int perimeter = m_sideLength * m_sides;
cout << "Square perimeter length: " << perimeter << endl;
}
};
class Triangle : public Shape {
int m_sides{0};
int m_sideLength{0};
public:
Triangle() = delete; // suppress default constructor
Triangle(int sides, int sideLength): m_sides(sides), m_sideLength(sideLength) {}
~Triangle() {}
void calculatePerimeter() {
int perimeter = m_sideLength * m_sides;
cout << "Triangle perimeter length: " << perimeter << endl;
}
};
int main() {
vector<Shape*> vec;
vec.push_back(new Square(4, 7));
vec.push_back(new Triangle(3, 13));
for (auto x: vec) {
x->calculatePerimeter();
}
// clean up
purge(vec);
system("pause");
return 0;
}
模板方式:
template <std::size_t SideCount, typename T>
class EquilateralPolygon
{
public:
explicit EquilateralPolygon(T sideLength) : sideLength(sideLength) {}
std::size_t GetSideCount() const { return SideCount; }
T GetSideLength() const { return sideLength; }
T calculatePerimeter() const { return SideCount * sideLength; }
private:
T sideLength;
};
template <std::size_t SideCount, typename T>
void displayPerimeter(const EquilateralPolygon<SideCount, T>& p) {
std::cout << "perimeter length: " << p.calculatePerimeter() << std::endl;
}
using EquilateralTriangle = EquilateralPolygon<3, std::size_t>;
using Square = EquilateralPolygon<4, std::size_t>;
使用方法:
int main() {
Square s(5);
std::cout << "number of sides a square has: " << s.GetSideCount() << std::endl;
EquilateralTrianglet(5);
std::cout << "number of sides a triangle has: " << t.GetSideCount() << std::endl;
displayPerimeter(s);
displayPerimeter(t);
}
模板方式:
template <std::size_t SideCount, typename T>
class EquilateralPolygon
{
public:
explicit EquilateralPolygon(T sideLength) : sideLength(sideLength) {}
std::size_t GetSideCount() const { return SideCount; }
T GetSideLength() const { return sideLength; }
T calculatePerimeter() const { return SideCount * sideLength; }
private:
T sideLength;
};
template <std::size_t SideCount, typename T>
void displayPerimeter(const EquilateralPolygon<SideCount, T>& p) {
std::cout << "perimeter length: " << p.calculatePerimeter() << std::endl;
}
using EquilateralTriangle = EquilateralPolygon<3, std::size_t>;
using Square = EquilateralPolygon<4, std::size_t>;
使用方法:
int main() {
Square s(5);
std::cout << "number of sides a square has: " << s.GetSideCount() << std::endl;
EquilateralTrianglet(5);
std::cout << "number of sides a triangle has: " << t.GetSideCount() << std::endl;
displayPerimeter(s);
displayPerimeter(t);
}
如果删除int,则会发生编译时错误。这是如何解决上述问题的?为每个被重写的类编写一个构造函数,并将变量移到其中。类三角形:public Shape{public:Triangle{sides=3;sideLength=4;};void calculatePerimeter{int permiture=sideLength*sides;std::cout您应该将其与您的答案结合起来,使其成为一个完整的答案。如果删除int,则会发生编译时错误。这如何解决上述问题?为每个重写的类编写构造函数,并将变量移到其中。类三角形:公共形状{public:Triangle{sides=3;sideLength=4;};void calculatePerimeter{int perime=si
删除*边的长度;std::你能不能把这个和你的答案结合起来,使它成为一个完整的答案。但是正方形总是有4条边,所以它不应该是一个从外部给定的变量。但是正方形总是有4条边,所以它不应该是一个从外部给定的变量。所以我们可以创建42条边的正方形或三角形:-所以我们可以创建42条边的正方形或三角形:-注意该Square实际上有4个成员:Shape::sides,Shape::sideLength,Square::sideLength…@Jarod42确实如此,我来自Java,在扩展类中声明一个变量会覆盖该变量,而不是复制它…看起来我有很多东西要学。注意Square实际上有4个成员:Shape::sides,Shape::sideLength,Square::sideLength,Square::sideLength…@Jarod42确实如此,我来自Java,在扩展类中声明变量会覆盖变量,而不是复制变量…看起来我有很多东西要学。没有解释的代码不是一个好答案,特别是如果代码没有回答问题的话。他问了一些减少复制/粘贴代码的方法,但如果有一个非常简单的方法,你的答案仍然需要。@JohnDrouhard这是解决问题的另一种方法,是的,我的代码重复了函数定义,但至少没有重复其中的内容。此外,我想在这里指出调用基类方法的另一种方法。正如你所看到的,用户是一个“新手”,我不明白你怎么能在没有适当解释的情况下抛出包含“虚拟”和“模板”的答案,并期望用户理解?你说了一个简单的解决方案。我相信我的代码是上面所有方法中最简单的。请告诉我你认为哪种解决方案对“新手”来说是最简单的。没有解释的代码不是一个好答案,特别是如果代码没有回答问题。他问了一些减少复制/粘贴代码的方法,但如果有一个非常简单的方法,你的答案仍然需要。@JohnDrouhard这是解决问题的另一种方法,是的,我的代码重复了函数定义,但至少没有重复其中的内容。此外,我想在这里指出调用基类方法的另一种方法。正如你所看到的,用户是一个“新手”,我不明白你怎么能在没有适当解释的情况下抛出包含“虚拟”和“模板”的答案,并期望用户理解?你说了一个简单的解决方案。我相信我的代码是上面所有方法中最简单的。请告诉我,根据您的说法,对于“新手”来说,哪种解决方案最简单。非常感谢您,它非常有效,而且远没有那么难看。谢谢你也解释了答案,我很感激。非常感谢你,它很好用,而且不那么难看。谢谢你也解释了答案,我很感激。