与操作符不匹配+ c++ 中的错误

Question

这是我一直在研究的 Rational 类：

rational.hrational.cpp

#include<iostream>

using namespace std;

#ifndef RATIONAL_H
#define RATIONAL_H

class Rational
{
  int numerator,denominator;
  public:
  // the various constructors
  Rational();
  Rational(int);
  Rational(int,int);

  //member functions
  int get_numerator()const{return numerator;}
  int get_denominator()const{return denominator;}

  // overloaded operators
  // relational operators
  bool operator==(const Rational&)const;
  bool operator<(const Rational&)const;
  bool operator<=(const Rational&)const;
  bool operator>(const Rational&)const;
  bool operator>=(const Rational&)const;

  //arithmetic operators
  Rational operator+(const Rational&);
  Rational operator-(const Rational&);
  Rational operator*(const Rational&);
  Rational operator/(const Rational&);

  //output operator
  friend ostream& operator<<(ostream&, const Rational&);
};
#endif //RATIONAL_H

这

#include "rational.h"

// implementation ofthe various constructors
Rational::Rational()
  :numerator(0),denominator(1){}

Rational::Rational(int number)
  :numerator(number),denominator(1){}

Rational::Rational(int n,int d)
  :numerator(n),denominator(d)
{
  if(denominator == 0) denominator = 1;
  if(denominator < 0)
  {
    numerator *= -1;
    denominator *= -1;
  }
}

// implementation of overloaded operators
bool Rational::operator==(const Rational& rhs)const
{
  if( numerator * rhs.get_denominator() == denominator * rhs.get_numerator() )
  {
    return true;
  }
  else return false;
}

bool Rational::operator<(const Rational& rhs)const
{
  if( numerator * rhs.get_denominator() < denominator * rhs.get_numerator() )
  {
    return true;
  }
  else return false;
}

bool Rational::operator<=(const Rational& rhs)const
{
  return operator==(rhs) || operator<(rhs);
}

bool Rational::operator>(const Rational& rhs)const
{
  return !operator<(rhs);
}

bool Rational::operator>=(const Rational& rhs)const
{
  return operator==(rhs) || operator>(rhs);
}

//arithmetic operators
Rational Rational::operator+(const Rational& rhs)
{
  return Rational( (numerator * rhs.get_denominator() + denominator*rhs.get_numerator()), (denominator * rhs.get_denominator()) );
}

Rational Rational::operator-(const Rational& rhs)
{
  //reuse of the + operator for substraction
  return operator+(Rational(-1*rhs.get_numerator(),rhs.get_denominator()));
}

Rational Rational::operator*(const Rational& rhs)
{
  return Rational(numerator * rhs.get_numerator(), denominator * rhs.get_denominator());
}

Rational Rational::operator/(const Rational& rhs)
{
  //reuse of the * operator as division is the inverse of multiplication
  return operator*(Rational(rhs.get_denominator(),rhs.get_numerator()));
}

// friend output operator
ostream& operator<<(ostream& os, const Rational& r)
{
   os<<r.get_numerator()<<"/"<<r.get_denominator();
   return os;
}

and the driver for the program driver.cpp
#include "rational.h"

int main()
{
  Rational r1(),r2(3),r3(11,3),tmp;
  cout<<r1+r2<<endl;
  cout<<r2<<endl;
  cout<<r2-r3<<endl;
  cout<<r2*r3<<endl;
  cout<<r1/r3;

  return 0;
}

是我在尝试编译它时遇到的错误。

driver.cpp: In function ‘int main()’:
driver.cpp:6:12: error: no match for ‘operator+’ in ‘r1 + r2’
driver.cpp:6:12: note: candidates are:
/usr/include/c++/4.6/bits/stl_iterator.h:327:5: note: template<class _Iterator> std::reverse_iterator<_Iterator> std::operator+(typename std::reverse_iterator<_Iterator>::difference_type, const std::reverse_iterator<_Iterator>&)
/usr/include/c++/4.6/bits/basic_string.h:2306:5: note: template<class _CharT, class _Traits, class _Alloc> std::basic_string<_CharT, _Traits, _Alloc> std::operator+(const std::basic_string<_CharT, _Traits, _Alloc>&, const std::basic_string<_CharT, _Traits, _Alloc>&)
/usr/include/c++/4.6/bits/basic_string.tcc:694:5: note: template<class _CharT, class _Traits, class _Alloc> std::basic_string<_CharT, _Traits, _Alloc> std::operator+(const _CharT*, const std::basic_string<_CharT, _Traits, _Alloc>&)
/usr/include/c++/4.6/bits/basic_string.tcc:710:5: note: template<class _CharT, class _Traits, class _Alloc> std::basic_string<_CharT, _Traits, _Alloc> std::operator+(_CharT, const std::basic_string<_CharT, _Traits, _Alloc>&)
/usr/include/c++/4.6/bits/basic_string.h:2343:5: note: template<class _CharT, class _Traits, class _Alloc> std::basic_string<_CharT, _Traits, _Alloc> std::operator+(const std::basic_string<_CharT, _Traits, _Alloc>&, const _CharT*)
/usr/include/c++/4.6/bits/basic_string.h:2359:5: note: template<class _CharT, class _Traits, class _Alloc> std::basic_string<_CharT, _Traits, _Alloc> std::operator+(const std::basic_string<_CharT, _Traits, _Alloc>&, _CharT)
driver.cpp:10:12: error: no match for ‘operator/’ in ‘r1 / r3’

当我注释使用运算符+和运算符/的行时，代码就可以工作。 这让我感到困惑，因为我已经使用operator+实现了operator-，并且类似地使用operator*实现了operator/。 因此，如果其中一个有效，我认为另一个也会有效。 有人可以解释一下我在这里做错了什么吗？

编辑 当我使用运算符 == 时，我会收到更多错误

rational.cpp:22:6: error: prototype for ‘bool Rational::operator==(const Rational&)’ does not match any in class ‘Rational’
rational.h:23:8: error: candidate is: bool Rational::operator==(Rational)
rational.cpp:31:6: error: prototype for ‘bool Rational::operator<(const Rational&) const’ does not match any in class ‘Rational’
rational.h:24:8: error: candidate is: bool Rational::operator<(Rational)
driver.cpp: In function ‘int main()’:
driver.cpp:11:10: error: no match for ‘operator==’ in ‘r1 == tmp’
driver.cpp:11:10: note: candidates are:
/usr/include/c++/4.6/bits/postypes.h:218:5: note: template<class _StateT> bool std::operator==(const std::fpos<_StateT>&, const std::fpos<_StateT>&)
/usr/include/c++/4.6/bits/stl_pair.h:201:5: note: template<class _T1, class _T2> bool std::operator==(const std::pair<_T1, _T2>&, const std::pair<_T1, _T2>&)
/usr/include/c++/4.6/bits/stl_iterator.h:285:5: note: template<class _Iterator> bool std::operator==(const std::reverse_iterator<_Iterator>&, const std::reverse_iterator<_Iterator>&)
/usr/include/c++/4.6/bits/stl_iterator.h:335:5: note: template<class _IteratorL, class _IteratorR> bool std::operator==(const std::reverse_iterator<_IteratorL>&, const std::reverse_iterator<_IteratorR>&)
/usr/include/c++/4.6/bits/allocator.h:122:5: note: template<class _T1, class _T2> bool std::operator==(const std::allocator<_T1>&, const std::allocator<_T2>&)
/usr/include/c++/4.6/bits/allocator.h:127:5: note: template<class _Tp> bool std::operator==(const std::allocator<_Tp1>&, const std::allocator<_Tp1>&)
/usr/include/c++/4.6/bits/basic_string.h:2427:5: note: template<class _CharT, class _Traits, class _Alloc> bool std::operator==(const std::basic_string<_CharT, _Traits, _Alloc>&, const std::basic_string<_CharT, _Traits, _Alloc>&)
/usr/include/c++/4.6/bits/basic_string.h:2434:5: note: template<class _CharT> typename __gnu_cxx::__enable_if<std::__is_char<_Tp>::__value, bool>::__type std::operator==(const std::basic_string<_CharT>&, const std::basic_string<_CharT>&)
/usr/include/c++/4.6/bits/basic_string.h:2448:5: note: template<class _CharT, class _Traits, class _Alloc> bool std::operator==(const _CharT*, const std::basic_string<_CharT, _Traits, _Alloc>&)
/usr/include/c++/4.6/bits/basic_string.h:2460:5: note: template<class _CharT, class _Traits, class _Alloc> bool std::operator==(const std::basic_string<_CharT, _Traits, _Alloc>&, const _CharT*)
/usr/include/c++/4.6/bits/streambuf_iterator.h:194:5: note: template<class _CharT, class _Traits> bool std::operator==(const std::istreambuf_iterator<_CharT, _Traits>&, const std::istreambuf_iterator<_CharT, _Traits>&)

这些神秘消息意味着什么？

Answer 1

r1 不是 Rational 类型的对象，正如您所声明的那样，它是一个不带参数并返回 Rational 的函数。

r1() + r2 将是一个有效的表达式，但您可能想让 r1 成为 Rational：

Rational r1,r2(3),r3(11,3),tmp;

请注意，您的 operator+< /code> 相对于 const 是不对称的。我通常建议将 operator+ 设为自由函数：

Rational operator+( const Rational&, const Rational& );

如果您希望它成为成员函数，则可能应该将其声明为 const：

Rational operator+( const Rational& ) const;