Question

我正在努力添加表示适当数学对象的类Natural，Rational，Complex的操作。我需要用x来计算多项式。

所有类都继承抽象类Number。将所有系数都放在数组中我想计算多项式。为此，我需要乘以double的运算（x是double）。 x被转换为Rational并成倍增加。这很好用。我的问题是如何添加抽象类型Number？

的类

我不能让它发挥作用。我得到的就是永远不会在Number :: add（Number）中结束递归（它调用自身而不是为类型Natural，Rational，Complex调用其他方法）。

＃包括＃包括＃包括＃包括＃包括＃包括＃包括使用namespace std;

class Natural;class Rational;class Complex;

class Number {
public:
  virtual string toString() const = 0;
  virtual Number *operator*(const Rational) const = 0;
  virtual Number *add(const Natural*) const = 0;
  virtual Number *add(const Rational*) const = 0;
  virtual Number *add(const Complex*) const = 0;
  virtual Number *add(const Number *n) const {
    n->add(this);
  }
};

class Natural : public Number {
  friend class Complex;
  int n;
public:
  Natural(const Natural &s) {
    n = s.n;
  }
  Natural(int number) : n(number) {}
  string toString() const {
    stringstream ss;
    ss << n;
    return ss.str();
  }
  operator Rational() const;
  operator Complex() const;
  operator int() const {
    return n;
  }
  Number *operator*(const Rational r) const;
  Number *add(const Natural* number) const {
    return new Natural(n + number->n);
  }
  Number *add(const Rational*) const;
  Number *add(const Complex*) const;
};

class Rational : public Number {
  friend class Natural;
  int numerator, denominator;
  void divideByGCD() {
    int a = numerator, b = denominator;
    //cout << a << ' ' << b << ' ';
    if(a < b) {
      int temp = a;
      a = b;
      b = temp;
    }
    while (b > 0) {
      int r = a % b;
      a = b; b = r;
      //cout << r << endl;
    }
    numerator /= a;
    denominator /= a;
    //cout << a << endl;
  }
public:
  Rational() {}
  Rational(const Rational &s) {
    numerator = s.numerator;
    denominator = s.denominator;
  }
  Rational(int n, int d) {
    if(d == 0) throw new runtime_error("denominator equals 0");
    if(d < 0) {
      numerator = -n;
      denominator = -d;
    } else {
      numerator = n;
      denominator = d;
    }
    divideByGCD();
  }
  Rational(double d) {
    int i = 0, mul = 1;
    int r = d-floor(d);;
    while(r!=0) {
      i++; mul *= 10;
      r = 10*r-floor(10*r);
    }
    numerator = (int)mul*d;
    denominator = mul;
    divideByGCD();
  }
  string toString() const {
    stringstream ss;
    ss << numerator;
    if(denominator > 1) ss << '/' << denominator;
    return ss.str();
  }
  operator const Complex() const;
  operator const double() const {
    return (double)numerator/denominator;
  }
  Number *operator*(const Rational r) const {
    return new Rational(numerator*r.numerator, denominator*r.denominator);
  }
  Number *add(const Rational* r) const {
    return new Rational(numerator*r->denominator+r->numerator*denominator, denominator*r->denominator);
  }
  Number *add(const Natural*) const;
  Number *add(const Complex*) const;
};

class Complex : public Number {
  friend class Rational;
  double real, imaginary;
  static const double radius = 10;
public:
  Complex() {}
  Complex(const Complex &s) {
    real = s.real;
    imaginary = s.imaginary;
  }
  Complex(const double r, const double im) : real(r), imaginary(im) {}
  string toString() const {
    stringstream ss;
    ss << real;
    if(imaginary != 0) ss << '+' << imaginary << 'i';
    return ss.str();
  }
  Number *operator*(const Rational r) const;
  Number *add(const Complex* c) const {
    return new Complex(real + c->real, imaginary + c->imaginary);
  }
  Number *add(const Natural*) const;
  Number *add(const Rational*) const;
};

Natural::operator Rational() const {
  return Rational(n,1);
}
Natural::operator Complex() const {
  return Complex(n, 0);
}
Rational::operator const Complex() const {
  return Complex((double)numerator/denominator, 0);
}

Number *Natural::operator*(const Rational r) const {
  return new Rational(n*r.numerator, r.denominator);
}
Number *Complex::operator*(const Rational r) const {
  return new Complex(real*(double)r, imaginary*(double)r);
}

Number *Natural::add(const Rational *r) const {
  if(r->denominator == 1) return new Natural(n+r->numerator);
  else return new Rational(n*r->denominator,r->denominator);
}

Number *Natural::add(const Complex *c) const {
  return c->add(this);
}

Number *Rational::add(const Natural *n) const {
  return n->add(this);
}

Number *Rational::add(const Complex *c) const {
  return new Complex(c->real+(double)*this, c->imaginary);
}

Number *Complex::add(const Natural *number) const {
  return new Complex(real+number->n, imaginary);
}

Number *Complex::add(const Rational *r) const {
  return r->add(this);
}

Number *poly(double x, Number *a[], unsigned int size) {
  if(size == 1) return a[0];
  else return a[0]->add((*poly(x, a+1, size-1))*Rational(x));
}

int main() {
  cout << (Natural(5)*(Rational)2.0)->toString() << endl;

  Number *coefs[] = {new Natural(5), new Natural(6)};
  cout <<  poly(2, coefs, 2) << endl;
}

我应该如何修复Number :: add（Number）以便在调用添加类型为Number的对象时自动确定添加哪个虚拟方法？

Answer 1

我认为问题是：

virtual Number *add(const Number *n) const {
   n->add(this);
}

如果将Rational乘以存储在Number *中的Natural，则不能将Number *多态转换为Natural *。我同意w / g-makulik在这里引用/值更有意义，因为你在整个地方泄漏了内存。删除对Number + Number的支持。另外，如果我一起添加一个Natural和一个Rational，我会得到一个Number *，但它是什么类型的数字？我认为架构需要更多思考;我可能完全摆脱基类纯虚方法（除了toString）。例如：

class Number
{
    public:
        virtual string toString() = 0;
};

class Rational : public Number
{
    string toString() {...}
    // forget 'add', use operators
    Rational operator+(const Rational & _rhs) const {Rational ret; ...; return ret;}
    Rational & operator+=(const Rational & _rhs) const {...; return *this;}
    ...
}

修改为了快速解决问题，我认为您只需摆脱virtual Number *operator*(const Rational) const = 0;，并将其替换为每个子类的版本（e.x。，Rational * operator*(const Natural) const）

或者，将枚举成员变量添加到Number以跟踪类型：

enum Type { NATURAL, RATIONAL, ...} Type mType;

或使用RTTI，这样您就可以在Number :: add：
中有选择地选择正确的add方法
Number * add(Number * _rhs) { if(_rhs->mType == RATIONAL) return this->add((Rational *)_rhs); ... }
它看起来有点草率，但它会起作用

Answer 2

这称为多次发送。以下是一些要查看的链接

Multiple_dispatch

best multimethods implementation

Answer 3

看起来Visitor pattern就像我一直在寻找的那样。我希望函数接受并访问同一个类。我相信我的错误就是给他们相同的名字。

用系数计算多项式作为抽象类数

3 个答案: