c++ - 在 C++ 类中具有公共(public)属性

Question

我如何在 C++ 类中拥有属性，就像在 C# 类中一样。

我不想拥有 getter 和 setter 方法。

Answer 1

您可以使用类似于 Jon 建议的解决方案，但使用运算符重载保留普通 C++ 语义。我稍微修改了 Jon 的代码如下(代码后面有解释):

#include <iostream>

template<typename T>
class Accessor {
public:
    explicit Accessor(const T& data) : value(data) {}

    Accessor& operator=(const T& data) { value = data; return *this; }
    Accessor& operator=(const Accessor& other) { this->value = other.value; return *this; }
    operator T() const { return value; }
    operator T&() { return value; }

private:
    Accessor(const Accessor&);


    T value;

};

struct Point {
    Point(int a = 0, int b = 0) : x(a), y(b) {}
    Accessor<int> x;
    Accessor<int> y;
};

int main() {
    Point p;
    p.x = 10;
    p.y = 20;
    p.x++;
    std::cout << p.x << "," << p.y << std::endl;

    p.x = p.y = 15;
    std::cout << p.x << "," << p.y << std::endl;

    return 0;
}

我们重载 operator= 以保留通常的赋值语法，而不是类似函数调用的语法。我们使用强制转换运算符作为“getter”。我们需要第二个版本的 operator= 来允许在 main() 中进行第二种赋值。

现在您可以添加到访问器的构造函数指针，或者更好的 - 仿函数 - 以任何方式调用 getter/setter 对您来说似乎是正确的。下面的例子假设 setter 函数返回 bool 表示同意设置新值，getter 可以在它退出时修改它:

#include <iostream>
#include <functional>
#include <cmath>

template<typename T>
class MySetter {
public:
    bool operator()(const T& data)
    {
        return (data <= 20 ? true : false);
    }
};

template<typename T>
class MyGetter {
public:
    T operator()(const T& data)
    {
        return round(data, 2);
    }

private:
    double cint(double x) {
        double dummy;
        if (modf(x,&dummy) >= 0.5) {
            return (x >= 0 ? ceil(x) : floor(x));
        } else {
            return (x < 0 ? ceil(x) : floor(x));
        }
    }

    double round(double r, int places) {
        double off = pow(10.0L, places);
        return cint(r*off)/off;
    }
};

template<typename T, typename G = MyGetter<T>, typename S = MySetter<T>>
class Accessor {
public:
    explicit Accessor(const T& data, const G& g = G(), const S& s = S()) : value(data), getter(g), setter(s) {}

    Accessor& operator=(const T& data) { if (setter(data)) value = data; return *this; }
    Accessor& operator=(const Accessor& other) { if (setter(other.value)) this->value = other.value; return *this; }
    operator T() const { value = getter(value); return value;}
    operator T&() { value = getter(value); return value; }

private:
    Accessor(const Accessor&);

    T value;

    G getter;
    S setter;

};

struct Point {
    Point(double a = 0, double b = 0) : x(a), y(b) {}
    Accessor<double> x;
    Accessor<double> y;
};

int main() {
    Point p;
    p.x = 10.712;
    p.y = 20.3456;
    p.x+=1;
    std::cout << p.x << "," << p.y << std::endl;

    p.x = p.y = 15.6426;
    std::cout << p.x << "," << p.y << std::endl;

    p.x = p.y = 25.85426;
    std::cout << p.x << "," << p.y << std::endl;

    p.x = p.y = 19.8425;
    p.y+=1;
    std::cout << p.x << "," << p.y << std::endl;

    return 0;
}

但是，正如最后一行所示，它有一个错误。返回 T& 的 cast 运算符允许用户绕过 setter，因为它使他们可以访问私有(private)值。解决此错误的一种方法是实现您希望访问器提供的所有运算符。例如，在以下代码中，我使用了 += 运算符，并且由于我删除了强制转换运算符返回引用，因此我必须实现一个 operator+=:

#include <iostream>
#include <functional>
#include <cmath>

template<typename T>
class MySetter {
public:
    bool operator()(const T& data) const {
        return (data <= 20 ? true : false);
    }
};

template<typename T>
class MyGetter {
public:
    T operator() (const T& data) const {
        return round(data, 2);
    }

private:
    double cint(double x) const {
        double dummy;
        if (modf(x,&dummy) >= 0.5) {
            return (x >= 0 ? ceil(x) : floor(x));
        } else {
            return (x < 0 ? ceil(x) : floor(x));
        }
    }

    double round(double r, int places) const {
        double off = pow(10.0L, places);
        return cint(r*off)/off;
    }
};

template<typename T, typename G = MyGetter<T>, typename S = MySetter<T>>
class Accessor {
private:
public:
    explicit Accessor(const T& data, const G& g = G(), const S& s = S()) : value(data), getter(g), setter(s) {}

    Accessor& operator=(const T& data) { if (setter(data)) value = data; return *this; }
    Accessor& operator=(const Accessor& other) { if (setter(other.value)) this->value = other.value; return *this; }
    operator T() const { return getter(value);}

    Accessor& operator+=(const T& data) { if (setter(value+data)) value += data; return *this; }

private:
    Accessor(const Accessor&);

    T value;

    G getter;
    S setter;

};

struct Point {
    Point(double a = 0, double b = 0) : x(a), y(b) {}
    Accessor<double> x;
    Accessor<double> y;
};

int main() {
    Point p;
    p.x = 10.712;
    p.y = 20.3456;
    p.x+=1;
    std::cout << p.x << "," << p.y << std::endl;

    p.x = p.y = 15.6426;
    std::cout << p.x << "," << p.y << std::endl;

    p.x = p.y = 25.85426;
    std::cout << p.x << "," << p.y << std::endl;

    p.x = p.y = 19.8425;
    p.y+=1;
    std::cout << p.x << "," << p.y << std::endl;

    return 0;
}

您必须实现所有将要使用的运算符。