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Template code - and is this redundancy or intended?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

-1

$begingroup$

I would like to discuss this code which is used in a PDE solver framework I found online. I don't quiet understand why I would need to introduce this form of code duplication.

template <std::size_t N, typename T>
class MathVector

 public:
 typedef T value_type;
 typedef std::size_t size_type;
 static const std::size_t Size = N;

 public:
 MathVector() 
 MathVector(const value_type& val) for(std::size_t i = 0; i < N; ++i) m_data[i] = val;


 // negation
 MathVector& operator- () for(std::size_t i = 0; i < N; ++i) m_data[i] = -m_data[i]; return *this;

Then there is classes for dimension 1 and 2...

template <typename T>
class MathVector<1, T>

 public:
 typedef std::size_t size_type;
 typedef T value_type;
 static const std::size_t Size = 1;

 public:
 MathVector() 
 MathVector(value_type x)
 
 m_data[0] = x;
 

 // negation
 MathVector& operator- () m_data[0] = -m_data[0]; return *this;



template <typename T>
class MathVector<2, T>

 public:
 typedef std::size_t size_type;
 typedef T value_type;
 static const std::size_t Size = 2;

 public:
 MathVector() 
 MathVector(value_type x, value_type y)
 
 m_data[0] = x;
 m_data[1] = y;
 

 // negation
 MathVector& operator- () m_data[0] = -m_data[0]; m_data[1] = -m_data[1]; return *this;

and for 3 and 4 analogue to that. I'm wondering why I would duplicate this. Wouldn't the first template class be sufficient?

c++ template

share|improve this question

edited 5 mins ago

stephanmg

asked 13 mins ago

stephanmgstephanmg

12

New contributor

stephanmg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sufficient for what?
  $endgroup$
  – Juho
  12 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, the missing code was not saved.
  $endgroup$
  – stephanmg
  10 mins ago
  

  
  
  
  

add a comment | 

-1

$begingroup$

I would like to discuss this code which is used in a PDE solver framework I found online. I don't quiet understand why I would need to introduce this form of code duplication.

template <std::size_t N, typename T>
class MathVector

 public:
 typedef T value_type;
 typedef std::size_t size_type;
 static const std::size_t Size = N;

 public:
 MathVector() 
 MathVector(const value_type& val) for(std::size_t i = 0; i < N; ++i) m_data[i] = val;


 // negation
 MathVector& operator- () for(std::size_t i = 0; i < N; ++i) m_data[i] = -m_data[i]; return *this;

Then there is classes for dimension 1 and 2...

template <typename T>
class MathVector<1, T>

 public:
 typedef std::size_t size_type;
 typedef T value_type;
 static const std::size_t Size = 1;

 public:
 MathVector() 
 MathVector(value_type x)
 
 m_data[0] = x;
 

 // negation
 MathVector& operator- () m_data[0] = -m_data[0]; return *this;



template <typename T>
class MathVector<2, T>

 public:
 typedef std::size_t size_type;
 typedef T value_type;
 static const std::size_t Size = 2;

 public:
 MathVector() 
 MathVector(value_type x, value_type y)
 
 m_data[0] = x;
 m_data[1] = y;
 

 // negation
 MathVector& operator- () m_data[0] = -m_data[0]; m_data[1] = -m_data[1]; return *this;

and for 3 and 4 analogue to that. I'm wondering why I would duplicate this. Wouldn't the first template class be sufficient?

c++ template

share|improve this question

edited 5 mins ago

stephanmg

asked 13 mins ago

stephanmgstephanmg

12

New contributor

stephanmg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sufficient for what?
  $endgroup$
  – Juho
  12 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, the missing code was not saved.
  $endgroup$
  – stephanmg
  10 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

I would like to discuss this code which is used in a PDE solver framework I found online. I don't quiet understand why I would need to introduce this form of code duplication.

template <std::size_t N, typename T>
class MathVector

 public:
 typedef T value_type;
 typedef std::size_t size_type;
 static const std::size_t Size = N;

 public:
 MathVector() 
 MathVector(const value_type& val) for(std::size_t i = 0; i < N; ++i) m_data[i] = val;


 // negation
 MathVector& operator- () for(std::size_t i = 0; i < N; ++i) m_data[i] = -m_data[i]; return *this;

Then there is classes for dimension 1 and 2...

template <typename T>
class MathVector<1, T>

 public:
 typedef std::size_t size_type;
 typedef T value_type;
 static const std::size_t Size = 1;

 public:
 MathVector() 
 MathVector(value_type x)
 
 m_data[0] = x;
 

 // negation
 MathVector& operator- () m_data[0] = -m_data[0]; return *this;



template <typename T>
class MathVector<2, T>

 public:
 typedef std::size_t size_type;
 typedef T value_type;
 static const std::size_t Size = 2;

 public:
 MathVector() 
 MathVector(value_type x, value_type y)
 
 m_data[0] = x;
 m_data[1] = y;
 

 // negation
 MathVector& operator- () m_data[0] = -m_data[0]; m_data[1] = -m_data[1]; return *this;

and for 3 and 4 analogue to that. I'm wondering why I would duplicate this. Wouldn't the first template class be sufficient?

c++ template

share|improve this question

edited 5 mins ago

stephanmg

asked 13 mins ago

stephanmgstephanmg

12

New contributor

stephanmg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

template <std::size_t N, typename T>
class MathVector

 public:
 typedef T value_type;
 typedef std::size_t size_type;
 static const std::size_t Size = N;

 public:
 MathVector() 
 MathVector(const value_type& val) for(std::size_t i = 0; i < N; ++i) m_data[i] = val;


 // negation
 MathVector& operator- () for(std::size_t i = 0; i < N; ++i) m_data[i] = -m_data[i]; return *this;

template <typename T>
class MathVector<1, T>

 public:
 typedef std::size_t size_type;
 typedef T value_type;
 static const std::size_t Size = 1;

 public:
 MathVector() 
 MathVector(value_type x)
 
 m_data[0] = x;
 

 // negation
 MathVector& operator- () m_data[0] = -m_data[0]; return *this;



template <typename T>
class MathVector<2, T>

 public:
 typedef std::size_t size_type;
 typedef T value_type;
 static const std::size_t Size = 2;

 public:
 MathVector() 
 MathVector(value_type x, value_type y)
 
 m_data[0] = x;
 m_data[1] = y;
 

 // negation
 MathVector& operator- () m_data[0] = -m_data[0]; m_data[1] = -m_data[1]; return *this;

share|improve this question

edited 5 mins ago

stephanmg

asked 13 mins ago

stephanmgstephanmg

12

New contributor

stephanmg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 5 mins ago

stephanmg

asked 13 mins ago

stephanmgstephanmg

12

New contributor

stephanmg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 13 mins ago

stephanmgstephanmg

12

New contributor

stephanmg is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 13 mins ago

stephanmgstephanmg

12