Template deduction complaints ambiguous candidates












6














I intend to implement the multiplication operator of my "Sparse Vector" and "Vector" classes. The following simplified code demo shows my problem



The Vector class in Vector.hpp



#pragma once

template <typename T>
class Vector
{
public:
Vector() {}

template <typename Scalar>
friend Vector operator*(const Scalar &a, const Vector &rhs) // #1
{
return Vector();
}
};


The Sparse Vector class in SpVec.hpp



#pragma once
#include "Vector.hpp"

template <typename T>
class SpVec
{
public:
SpVec() {}

template <typename U>
inline friend double operator*(const SpVec &spv, const Vector<U> &v) // #2
{
return 0.0;
}
};


The test code in main.cpp:



#include "Vector.hpp"
#include "SpVec.hpp"


#include <iostream>

int main()
{
Vector<double> v;

SpVec<double> spv;

std::cout << spv * v;
return 0;
}


I build the test program with



g++ main.cpp -o test


which gives the ambiguous template deduction error



main.cpp: In function ‘int main()’:
main.cpp:13:26: error: ambiguous overload for ‘operator*’ (operand types are ‘SpVec<double>’ and ‘Vector<double>’)
std::cout << spv * v;
~~~~^~~
In file included from main.cpp:2:0:
SpVec.hpp:12:26: note: candidate: double operator*(const SpVec<T>&, const Vector<U>&) [with U = double; T = double]
inline friend double operator*(const SpVec &spv, const Vector<U> &v) // #2
^~~~~~~~
In file included from main.cpp:1:0:
Vector.hpp:10:19: note: candidate: Vector<T> operator*(const Scalar&, const Vector<T>&) [with Scalar = SpVec<double>; T = double]
friend Vector operator*(const Scalar &a, const Vector &rhs) // #1


I expect the #2 method definition is more close to my calling.



Please help me understand how the ambiguous error comes out and how to resolve the issue.










share|improve this question



























    6














    I intend to implement the multiplication operator of my "Sparse Vector" and "Vector" classes. The following simplified code demo shows my problem



    The Vector class in Vector.hpp



    #pragma once

    template <typename T>
    class Vector
    {
    public:
    Vector() {}

    template <typename Scalar>
    friend Vector operator*(const Scalar &a, const Vector &rhs) // #1
    {
    return Vector();
    }
    };


    The Sparse Vector class in SpVec.hpp



    #pragma once
    #include "Vector.hpp"

    template <typename T>
    class SpVec
    {
    public:
    SpVec() {}

    template <typename U>
    inline friend double operator*(const SpVec &spv, const Vector<U> &v) // #2
    {
    return 0.0;
    }
    };


    The test code in main.cpp:



    #include "Vector.hpp"
    #include "SpVec.hpp"


    #include <iostream>

    int main()
    {
    Vector<double> v;

    SpVec<double> spv;

    std::cout << spv * v;
    return 0;
    }


    I build the test program with



    g++ main.cpp -o test


    which gives the ambiguous template deduction error



    main.cpp: In function ‘int main()’:
    main.cpp:13:26: error: ambiguous overload for ‘operator*’ (operand types are ‘SpVec<double>’ and ‘Vector<double>’)
    std::cout << spv * v;
    ~~~~^~~
    In file included from main.cpp:2:0:
    SpVec.hpp:12:26: note: candidate: double operator*(const SpVec<T>&, const Vector<U>&) [with U = double; T = double]
    inline friend double operator*(const SpVec &spv, const Vector<U> &v) // #2
    ^~~~~~~~
    In file included from main.cpp:1:0:
    Vector.hpp:10:19: note: candidate: Vector<T> operator*(const Scalar&, const Vector<T>&) [with Scalar = SpVec<double>; T = double]
    friend Vector operator*(const Scalar &a, const Vector &rhs) // #1


    I expect the #2 method definition is more close to my calling.



    Please help me understand how the ambiguous error comes out and how to resolve the issue.










    share|improve this question

























      6












      6








      6


      1





      I intend to implement the multiplication operator of my "Sparse Vector" and "Vector" classes. The following simplified code demo shows my problem



      The Vector class in Vector.hpp



      #pragma once

      template <typename T>
      class Vector
      {
      public:
      Vector() {}

      template <typename Scalar>
      friend Vector operator*(const Scalar &a, const Vector &rhs) // #1
      {
      return Vector();
      }
      };


      The Sparse Vector class in SpVec.hpp



      #pragma once
      #include "Vector.hpp"

      template <typename T>
      class SpVec
      {
      public:
      SpVec() {}

      template <typename U>
      inline friend double operator*(const SpVec &spv, const Vector<U> &v) // #2
      {
      return 0.0;
      }
      };


      The test code in main.cpp:



      #include "Vector.hpp"
      #include "SpVec.hpp"


      #include <iostream>

      int main()
      {
      Vector<double> v;

      SpVec<double> spv;

      std::cout << spv * v;
      return 0;
      }


      I build the test program with



      g++ main.cpp -o test


      which gives the ambiguous template deduction error



      main.cpp: In function ‘int main()’:
      main.cpp:13:26: error: ambiguous overload for ‘operator*’ (operand types are ‘SpVec<double>’ and ‘Vector<double>’)
      std::cout << spv * v;
      ~~~~^~~
      In file included from main.cpp:2:0:
      SpVec.hpp:12:26: note: candidate: double operator*(const SpVec<T>&, const Vector<U>&) [with U = double; T = double]
      inline friend double operator*(const SpVec &spv, const Vector<U> &v) // #2
      ^~~~~~~~
      In file included from main.cpp:1:0:
      Vector.hpp:10:19: note: candidate: Vector<T> operator*(const Scalar&, const Vector<T>&) [with Scalar = SpVec<double>; T = double]
      friend Vector operator*(const Scalar &a, const Vector &rhs) // #1


      I expect the #2 method definition is more close to my calling.



      Please help me understand how the ambiguous error comes out and how to resolve the issue.










      share|improve this question













      I intend to implement the multiplication operator of my "Sparse Vector" and "Vector" classes. The following simplified code demo shows my problem



      The Vector class in Vector.hpp



      #pragma once

      template <typename T>
      class Vector
      {
      public:
      Vector() {}

      template <typename Scalar>
      friend Vector operator*(const Scalar &a, const Vector &rhs) // #1
      {
      return Vector();
      }
      };


      The Sparse Vector class in SpVec.hpp



      #pragma once
      #include "Vector.hpp"

      template <typename T>
      class SpVec
      {
      public:
      SpVec() {}

      template <typename U>
      inline friend double operator*(const SpVec &spv, const Vector<U> &v) // #2
      {
      return 0.0;
      }
      };


      The test code in main.cpp:



      #include "Vector.hpp"
      #include "SpVec.hpp"


      #include <iostream>

      int main()
      {
      Vector<double> v;

      SpVec<double> spv;

      std::cout << spv * v;
      return 0;
      }


      I build the test program with



      g++ main.cpp -o test


      which gives the ambiguous template deduction error



      main.cpp: In function ‘int main()’:
      main.cpp:13:26: error: ambiguous overload for ‘operator*’ (operand types are ‘SpVec<double>’ and ‘Vector<double>’)
      std::cout << spv * v;
      ~~~~^~~
      In file included from main.cpp:2:0:
      SpVec.hpp:12:26: note: candidate: double operator*(const SpVec<T>&, const Vector<U>&) [with U = double; T = double]
      inline friend double operator*(const SpVec &spv, const Vector<U> &v) // #2
      ^~~~~~~~
      In file included from main.cpp:1:0:
      Vector.hpp:10:19: note: candidate: Vector<T> operator*(const Scalar&, const Vector<T>&) [with Scalar = SpVec<double>; T = double]
      friend Vector operator*(const Scalar &a, const Vector &rhs) // #1


      I expect the #2 method definition is more close to my calling.



      Please help me understand how the ambiguous error comes out and how to resolve the issue.







      c++ templates metaprogramming ambiguous






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 13 '18 at 5:58









      RubinRubin

      1037




      1037
























          2 Answers
          2






          active

          oldest

          votes


















          3














          I come up with another idea that the prior type information Scalar can be used with the SFAINE feature enabled by the c++11 standard library struct std::enable_if.



          The codes:



          Vector.hpp



          #pragma once

          #include <iostream>
          #include <type_traits>

          template <typename T>
          class Vector
          {
          public:
          Vector() {}

          template <typename Scalar>
          typename std::enable_if<std::is_arithmetic<Scalar>::value, Vector<T>>::type
          operator*(const Scalar &rhs) const// #1
          {
          std::cout << "Vector * Scalar called." << std::endl;
          return Vector();
          }

          template <typename Scalar>
          inline friend typename std::enable_if<std::is_arithmetic<Scalar>::value, Vector<T>>::type
          operator*(const Scalar &lhs, const Vector &rhs)
          {
          std::cout << "Scalar * Vector called." << std::endl;
          return Vector();
          }
          };


          SpVec.hpp



          #pragma once
          #include "Vector.hpp"

          #include <iostream>

          template <typename T>
          class SpVec
          {
          public:
          SpVec() {}

          template <typename U>
          inline double operator*(const Vector<U> &rhs) const // #2 as member function
          {
          std::cout << "SpVec * Vector called" << std::endl;
          return 0.0;
          }

          template <typename U>
          inline friend double operator*(const Vector<U> &lhs, const SpVec &rhs)
          {
          std::cout << "Vector * SpVec called" << std::endl;
          return 0.0;
          }
          };


          main.cpp



          #include "SpVec.hpp"
          #include "Vector.hpp"

          #include <iostream>

          int main()
          {
          Vector<double> v;
          SpVec<double> spv;

          double a = spv * v;
          a = v * spv;

          Vector<double> vt;
          vt = v * 2.0;
          vt = 2.0 * v;

          return 0;
          }


          Build the program with c++11



          g++ -std=c++11 main.cpp -o test


          The result:



          SpVec * Vector called.
          Vector * SpVec called.
          Vector * Scalar called.
          Scalar * Vector called.





          share|improve this answer























          • That's also a nice one !
            – Christophe
            Nov 13 '18 at 8:34



















          0














          The argument to operator* are SpVec<double> and Vector<double>. It could be resolved to



          operator*(const Scalar &a, const Vector &rhs) with scalar as SpVec<double> and rhs as Vector<double>.



          It could also resolve to



          operator*(const SpVec &spv, const Vector<U> &v) with spv as SpVec<double> and U as double.



          One way of resolving this is to turn Vector::operator* to non-friend function.



          Vector operator*(const Scalar &a)     // #1
          {
          //The other argument here will be accessed using this pointer.
          return Vector();
          }


          and you can call it as



          int main() 
          {
          Vector<double> v;
          SpVec<double> spv;
          std::cout << spv * v; // will call #2
          v * spv; //will call #1
          return 0;
          }





          share|improve this answer





















          • You beat me to it!
            – Francis Cugler
            Nov 13 '18 at 6:47










          • Thanks. This suggestion does work. For the deeper discussion, is it possible to make the two cases both work as no operands precedence need to be cared about for the user?
            – Rubin
            Nov 13 '18 at 6:57










          • Nice answer. However, this works only because you play with the order of the parameters (with an operator that is assumed to be commutative).
            – Christophe
            Nov 13 '18 at 7:06












          • @Rubin I think the term you are looking for is commutate and not precedence. The commutative property says that a*b is equal to b*a. Precedence is used to decide in which order several operators are executed: a + b * c where multiplication has a higher precedence and will be calculated before the addition. Also here a link to a question about commutative property in the non-templated case: stackoverflow.com/questions/3764604/…
            – Philipp
            Nov 13 '18 at 7:41










          • @Philipp Thanks for the correction. The commutative property is indeed what I need.
            – Rubin
            Nov 13 '18 at 7:49











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          2 Answers
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          active

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3














          I come up with another idea that the prior type information Scalar can be used with the SFAINE feature enabled by the c++11 standard library struct std::enable_if.



          The codes:



          Vector.hpp



          #pragma once

          #include <iostream>
          #include <type_traits>

          template <typename T>
          class Vector
          {
          public:
          Vector() {}

          template <typename Scalar>
          typename std::enable_if<std::is_arithmetic<Scalar>::value, Vector<T>>::type
          operator*(const Scalar &rhs) const// #1
          {
          std::cout << "Vector * Scalar called." << std::endl;
          return Vector();
          }

          template <typename Scalar>
          inline friend typename std::enable_if<std::is_arithmetic<Scalar>::value, Vector<T>>::type
          operator*(const Scalar &lhs, const Vector &rhs)
          {
          std::cout << "Scalar * Vector called." << std::endl;
          return Vector();
          }
          };


          SpVec.hpp



          #pragma once
          #include "Vector.hpp"

          #include <iostream>

          template <typename T>
          class SpVec
          {
          public:
          SpVec() {}

          template <typename U>
          inline double operator*(const Vector<U> &rhs) const // #2 as member function
          {
          std::cout << "SpVec * Vector called" << std::endl;
          return 0.0;
          }

          template <typename U>
          inline friend double operator*(const Vector<U> &lhs, const SpVec &rhs)
          {
          std::cout << "Vector * SpVec called" << std::endl;
          return 0.0;
          }
          };


          main.cpp



          #include "SpVec.hpp"
          #include "Vector.hpp"

          #include <iostream>

          int main()
          {
          Vector<double> v;
          SpVec<double> spv;

          double a = spv * v;
          a = v * spv;

          Vector<double> vt;
          vt = v * 2.0;
          vt = 2.0 * v;

          return 0;
          }


          Build the program with c++11



          g++ -std=c++11 main.cpp -o test


          The result:



          SpVec * Vector called.
          Vector * SpVec called.
          Vector * Scalar called.
          Scalar * Vector called.





          share|improve this answer























          • That's also a nice one !
            – Christophe
            Nov 13 '18 at 8:34
















          3














          I come up with another idea that the prior type information Scalar can be used with the SFAINE feature enabled by the c++11 standard library struct std::enable_if.



          The codes:



          Vector.hpp



          #pragma once

          #include <iostream>
          #include <type_traits>

          template <typename T>
          class Vector
          {
          public:
          Vector() {}

          template <typename Scalar>
          typename std::enable_if<std::is_arithmetic<Scalar>::value, Vector<T>>::type
          operator*(const Scalar &rhs) const// #1
          {
          std::cout << "Vector * Scalar called." << std::endl;
          return Vector();
          }

          template <typename Scalar>
          inline friend typename std::enable_if<std::is_arithmetic<Scalar>::value, Vector<T>>::type
          operator*(const Scalar &lhs, const Vector &rhs)
          {
          std::cout << "Scalar * Vector called." << std::endl;
          return Vector();
          }
          };


          SpVec.hpp



          #pragma once
          #include "Vector.hpp"

          #include <iostream>

          template <typename T>
          class SpVec
          {
          public:
          SpVec() {}

          template <typename U>
          inline double operator*(const Vector<U> &rhs) const // #2 as member function
          {
          std::cout << "SpVec * Vector called" << std::endl;
          return 0.0;
          }

          template <typename U>
          inline friend double operator*(const Vector<U> &lhs, const SpVec &rhs)
          {
          std::cout << "Vector * SpVec called" << std::endl;
          return 0.0;
          }
          };


          main.cpp



          #include "SpVec.hpp"
          #include "Vector.hpp"

          #include <iostream>

          int main()
          {
          Vector<double> v;
          SpVec<double> spv;

          double a = spv * v;
          a = v * spv;

          Vector<double> vt;
          vt = v * 2.0;
          vt = 2.0 * v;

          return 0;
          }


          Build the program with c++11



          g++ -std=c++11 main.cpp -o test


          The result:



          SpVec * Vector called.
          Vector * SpVec called.
          Vector * Scalar called.
          Scalar * Vector called.





          share|improve this answer























          • That's also a nice one !
            – Christophe
            Nov 13 '18 at 8:34














          3












          3








          3






          I come up with another idea that the prior type information Scalar can be used with the SFAINE feature enabled by the c++11 standard library struct std::enable_if.



          The codes:



          Vector.hpp



          #pragma once

          #include <iostream>
          #include <type_traits>

          template <typename T>
          class Vector
          {
          public:
          Vector() {}

          template <typename Scalar>
          typename std::enable_if<std::is_arithmetic<Scalar>::value, Vector<T>>::type
          operator*(const Scalar &rhs) const// #1
          {
          std::cout << "Vector * Scalar called." << std::endl;
          return Vector();
          }

          template <typename Scalar>
          inline friend typename std::enable_if<std::is_arithmetic<Scalar>::value, Vector<T>>::type
          operator*(const Scalar &lhs, const Vector &rhs)
          {
          std::cout << "Scalar * Vector called." << std::endl;
          return Vector();
          }
          };


          SpVec.hpp



          #pragma once
          #include "Vector.hpp"

          #include <iostream>

          template <typename T>
          class SpVec
          {
          public:
          SpVec() {}

          template <typename U>
          inline double operator*(const Vector<U> &rhs) const // #2 as member function
          {
          std::cout << "SpVec * Vector called" << std::endl;
          return 0.0;
          }

          template <typename U>
          inline friend double operator*(const Vector<U> &lhs, const SpVec &rhs)
          {
          std::cout << "Vector * SpVec called" << std::endl;
          return 0.0;
          }
          };


          main.cpp



          #include "SpVec.hpp"
          #include "Vector.hpp"

          #include <iostream>

          int main()
          {
          Vector<double> v;
          SpVec<double> spv;

          double a = spv * v;
          a = v * spv;

          Vector<double> vt;
          vt = v * 2.0;
          vt = 2.0 * v;

          return 0;
          }


          Build the program with c++11



          g++ -std=c++11 main.cpp -o test


          The result:



          SpVec * Vector called.
          Vector * SpVec called.
          Vector * Scalar called.
          Scalar * Vector called.





          share|improve this answer














          I come up with another idea that the prior type information Scalar can be used with the SFAINE feature enabled by the c++11 standard library struct std::enable_if.



          The codes:



          Vector.hpp



          #pragma once

          #include <iostream>
          #include <type_traits>

          template <typename T>
          class Vector
          {
          public:
          Vector() {}

          template <typename Scalar>
          typename std::enable_if<std::is_arithmetic<Scalar>::value, Vector<T>>::type
          operator*(const Scalar &rhs) const// #1
          {
          std::cout << "Vector * Scalar called." << std::endl;
          return Vector();
          }

          template <typename Scalar>
          inline friend typename std::enable_if<std::is_arithmetic<Scalar>::value, Vector<T>>::type
          operator*(const Scalar &lhs, const Vector &rhs)
          {
          std::cout << "Scalar * Vector called." << std::endl;
          return Vector();
          }
          };


          SpVec.hpp



          #pragma once
          #include "Vector.hpp"

          #include <iostream>

          template <typename T>
          class SpVec
          {
          public:
          SpVec() {}

          template <typename U>
          inline double operator*(const Vector<U> &rhs) const // #2 as member function
          {
          std::cout << "SpVec * Vector called" << std::endl;
          return 0.0;
          }

          template <typename U>
          inline friend double operator*(const Vector<U> &lhs, const SpVec &rhs)
          {
          std::cout << "Vector * SpVec called" << std::endl;
          return 0.0;
          }
          };


          main.cpp



          #include "SpVec.hpp"
          #include "Vector.hpp"

          #include <iostream>

          int main()
          {
          Vector<double> v;
          SpVec<double> spv;

          double a = spv * v;
          a = v * spv;

          Vector<double> vt;
          vt = v * 2.0;
          vt = 2.0 * v;

          return 0;
          }


          Build the program with c++11



          g++ -std=c++11 main.cpp -o test


          The result:



          SpVec * Vector called.
          Vector * SpVec called.
          Vector * Scalar called.
          Scalar * Vector called.






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 13 '18 at 11:14









          Oliv

          8,5191956




          8,5191956










          answered Nov 13 '18 at 8:30









          RubinRubin

          1037




          1037












          • That's also a nice one !
            – Christophe
            Nov 13 '18 at 8:34


















          • That's also a nice one !
            – Christophe
            Nov 13 '18 at 8:34
















          That's also a nice one !
          – Christophe
          Nov 13 '18 at 8:34




          That's also a nice one !
          – Christophe
          Nov 13 '18 at 8:34













          0














          The argument to operator* are SpVec<double> and Vector<double>. It could be resolved to



          operator*(const Scalar &a, const Vector &rhs) with scalar as SpVec<double> and rhs as Vector<double>.



          It could also resolve to



          operator*(const SpVec &spv, const Vector<U> &v) with spv as SpVec<double> and U as double.



          One way of resolving this is to turn Vector::operator* to non-friend function.



          Vector operator*(const Scalar &a)     // #1
          {
          //The other argument here will be accessed using this pointer.
          return Vector();
          }


          and you can call it as



          int main() 
          {
          Vector<double> v;
          SpVec<double> spv;
          std::cout << spv * v; // will call #2
          v * spv; //will call #1
          return 0;
          }





          share|improve this answer





















          • You beat me to it!
            – Francis Cugler
            Nov 13 '18 at 6:47










          • Thanks. This suggestion does work. For the deeper discussion, is it possible to make the two cases both work as no operands precedence need to be cared about for the user?
            – Rubin
            Nov 13 '18 at 6:57










          • Nice answer. However, this works only because you play with the order of the parameters (with an operator that is assumed to be commutative).
            – Christophe
            Nov 13 '18 at 7:06












          • @Rubin I think the term you are looking for is commutate and not precedence. The commutative property says that a*b is equal to b*a. Precedence is used to decide in which order several operators are executed: a + b * c where multiplication has a higher precedence and will be calculated before the addition. Also here a link to a question about commutative property in the non-templated case: stackoverflow.com/questions/3764604/…
            – Philipp
            Nov 13 '18 at 7:41










          • @Philipp Thanks for the correction. The commutative property is indeed what I need.
            – Rubin
            Nov 13 '18 at 7:49
















          0














          The argument to operator* are SpVec<double> and Vector<double>. It could be resolved to



          operator*(const Scalar &a, const Vector &rhs) with scalar as SpVec<double> and rhs as Vector<double>.



          It could also resolve to



          operator*(const SpVec &spv, const Vector<U> &v) with spv as SpVec<double> and U as double.



          One way of resolving this is to turn Vector::operator* to non-friend function.



          Vector operator*(const Scalar &a)     // #1
          {
          //The other argument here will be accessed using this pointer.
          return Vector();
          }


          and you can call it as



          int main() 
          {
          Vector<double> v;
          SpVec<double> spv;
          std::cout << spv * v; // will call #2
          v * spv; //will call #1
          return 0;
          }





          share|improve this answer





















          • You beat me to it!
            – Francis Cugler
            Nov 13 '18 at 6:47










          • Thanks. This suggestion does work. For the deeper discussion, is it possible to make the two cases both work as no operands precedence need to be cared about for the user?
            – Rubin
            Nov 13 '18 at 6:57










          • Nice answer. However, this works only because you play with the order of the parameters (with an operator that is assumed to be commutative).
            – Christophe
            Nov 13 '18 at 7:06












          • @Rubin I think the term you are looking for is commutate and not precedence. The commutative property says that a*b is equal to b*a. Precedence is used to decide in which order several operators are executed: a + b * c where multiplication has a higher precedence and will be calculated before the addition. Also here a link to a question about commutative property in the non-templated case: stackoverflow.com/questions/3764604/…
            – Philipp
            Nov 13 '18 at 7:41










          • @Philipp Thanks for the correction. The commutative property is indeed what I need.
            – Rubin
            Nov 13 '18 at 7:49














          0












          0








          0






          The argument to operator* are SpVec<double> and Vector<double>. It could be resolved to



          operator*(const Scalar &a, const Vector &rhs) with scalar as SpVec<double> and rhs as Vector<double>.



          It could also resolve to



          operator*(const SpVec &spv, const Vector<U> &v) with spv as SpVec<double> and U as double.



          One way of resolving this is to turn Vector::operator* to non-friend function.



          Vector operator*(const Scalar &a)     // #1
          {
          //The other argument here will be accessed using this pointer.
          return Vector();
          }


          and you can call it as



          int main() 
          {
          Vector<double> v;
          SpVec<double> spv;
          std::cout << spv * v; // will call #2
          v * spv; //will call #1
          return 0;
          }





          share|improve this answer












          The argument to operator* are SpVec<double> and Vector<double>. It could be resolved to



          operator*(const Scalar &a, const Vector &rhs) with scalar as SpVec<double> and rhs as Vector<double>.



          It could also resolve to



          operator*(const SpVec &spv, const Vector<U> &v) with spv as SpVec<double> and U as double.



          One way of resolving this is to turn Vector::operator* to non-friend function.



          Vector operator*(const Scalar &a)     // #1
          {
          //The other argument here will be accessed using this pointer.
          return Vector();
          }


          and you can call it as



          int main() 
          {
          Vector<double> v;
          SpVec<double> spv;
          std::cout << spv * v; // will call #2
          v * spv; //will call #1
          return 0;
          }






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 13 '18 at 6:40









          Gaurav SehgalGaurav Sehgal

          5,68111026




          5,68111026












          • You beat me to it!
            – Francis Cugler
            Nov 13 '18 at 6:47










          • Thanks. This suggestion does work. For the deeper discussion, is it possible to make the two cases both work as no operands precedence need to be cared about for the user?
            – Rubin
            Nov 13 '18 at 6:57










          • Nice answer. However, this works only because you play with the order of the parameters (with an operator that is assumed to be commutative).
            – Christophe
            Nov 13 '18 at 7:06












          • @Rubin I think the term you are looking for is commutate and not precedence. The commutative property says that a*b is equal to b*a. Precedence is used to decide in which order several operators are executed: a + b * c where multiplication has a higher precedence and will be calculated before the addition. Also here a link to a question about commutative property in the non-templated case: stackoverflow.com/questions/3764604/…
            – Philipp
            Nov 13 '18 at 7:41










          • @Philipp Thanks for the correction. The commutative property is indeed what I need.
            – Rubin
            Nov 13 '18 at 7:49


















          • You beat me to it!
            – Francis Cugler
            Nov 13 '18 at 6:47










          • Thanks. This suggestion does work. For the deeper discussion, is it possible to make the two cases both work as no operands precedence need to be cared about for the user?
            – Rubin
            Nov 13 '18 at 6:57










          • Nice answer. However, this works only because you play with the order of the parameters (with an operator that is assumed to be commutative).
            – Christophe
            Nov 13 '18 at 7:06












          • @Rubin I think the term you are looking for is commutate and not precedence. The commutative property says that a*b is equal to b*a. Precedence is used to decide in which order several operators are executed: a + b * c where multiplication has a higher precedence and will be calculated before the addition. Also here a link to a question about commutative property in the non-templated case: stackoverflow.com/questions/3764604/…
            – Philipp
            Nov 13 '18 at 7:41










          • @Philipp Thanks for the correction. The commutative property is indeed what I need.
            – Rubin
            Nov 13 '18 at 7:49
















          You beat me to it!
          – Francis Cugler
          Nov 13 '18 at 6:47




          You beat me to it!
          – Francis Cugler
          Nov 13 '18 at 6:47












          Thanks. This suggestion does work. For the deeper discussion, is it possible to make the two cases both work as no operands precedence need to be cared about for the user?
          – Rubin
          Nov 13 '18 at 6:57




          Thanks. This suggestion does work. For the deeper discussion, is it possible to make the two cases both work as no operands precedence need to be cared about for the user?
          – Rubin
          Nov 13 '18 at 6:57












          Nice answer. However, this works only because you play with the order of the parameters (with an operator that is assumed to be commutative).
          – Christophe
          Nov 13 '18 at 7:06






          Nice answer. However, this works only because you play with the order of the parameters (with an operator that is assumed to be commutative).
          – Christophe
          Nov 13 '18 at 7:06














          @Rubin I think the term you are looking for is commutate and not precedence. The commutative property says that a*b is equal to b*a. Precedence is used to decide in which order several operators are executed: a + b * c where multiplication has a higher precedence and will be calculated before the addition. Also here a link to a question about commutative property in the non-templated case: stackoverflow.com/questions/3764604/…
          – Philipp
          Nov 13 '18 at 7:41




          @Rubin I think the term you are looking for is commutate and not precedence. The commutative property says that a*b is equal to b*a. Precedence is used to decide in which order several operators are executed: a + b * c where multiplication has a higher precedence and will be calculated before the addition. Also here a link to a question about commutative property in the non-templated case: stackoverflow.com/questions/3764604/…
          – Philipp
          Nov 13 '18 at 7:41












          @Philipp Thanks for the correction. The commutative property is indeed what I need.
          – Rubin
          Nov 13 '18 at 7:49




          @Philipp Thanks for the correction. The commutative property is indeed what I need.
          – Rubin
          Nov 13 '18 at 7:49


















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