template template function definition

I have a template class called Array

template<typename T, int dimension>

class Array<typename T, int dimension>{



    //definition of the class



}

I want to write a non-member function cast such that I can cast Array into different type. For example

Array<int, 2> a;

Array<float, 2> b = cast<float>(a);

How should I write this function? I am more interested in how to declare it instead of how to implement the detailed casting. I have tried

template<template<typename T, int dimension> class Array, typename New_T, int dimension>

Array<typename New_T, int dimension> cast(Array<typename T, int dimension> a){



// detailed implementation of casting, which I do not care for this question.



}

but it cannot pass the compilation.

edited Nov 20 '18 at 18:19

max66

35.8k74165

asked Jun 27 '18 at 2:05

Nan

577

Without having tested it for myself, I'm not sure you actually need a template template here. I think the function can just be: template<typename T, int dimension, typename New_T> Array<New_T, dimension> cast(Array<T, dimension>)

– John Ilacqua
Jun 27 '18 at 2:18

add a comment |

I have a template class called Array

template<typename T, int dimension>

class Array<typename T, int dimension>{



    //definition of the class



}

I want to write a non-member function cast such that I can cast Array into different type. For example

Array<int, 2> a;

Array<float, 2> b = cast<float>(a);

How should I write this function? I am more interested in how to declare it instead of how to implement the detailed casting. I have tried

template<template<typename T, int dimension> class Array, typename New_T, int dimension>

Array<typename New_T, int dimension> cast(Array<typename T, int dimension> a){



// detailed implementation of casting, which I do not care for this question.



}

but it cannot pass the compilation.

edited Nov 20 '18 at 18:19

max66

35.8k74165

asked Jun 27 '18 at 2:05

Nan

577

Without having tested it for myself, I'm not sure you actually need a template template here. I think the function can just be: template<typename T, int dimension, typename New_T> Array<New_T, dimension> cast(Array<T, dimension>)

– John Ilacqua
Jun 27 '18 at 2:18

add a comment |

I have a template class called Array

template<typename T, int dimension>

class Array<typename T, int dimension>{



    //definition of the class



}

I want to write a non-member function cast such that I can cast Array into different type. For example

Array<int, 2> a;

Array<float, 2> b = cast<float>(a);

How should I write this function? I am more interested in how to declare it instead of how to implement the detailed casting. I have tried

template<template<typename T, int dimension> class Array, typename New_T, int dimension>

Array<typename New_T, int dimension> cast(Array<typename T, int dimension> a){



// detailed implementation of casting, which I do not care for this question.



}

but it cannot pass the compilation.

edited Nov 20 '18 at 18:19

max66

35.8k74165

asked Jun 27 '18 at 2:05

Nan

577

I have a template class called Array

template<typename T, int dimension>

class Array<typename T, int dimension>{



    //definition of the class



}

I want to write a non-member function cast such that I can cast Array into different type. For example

Array<int, 2> a;

Array<float, 2> b = cast<float>(a);

How should I write this function? I am more interested in how to declare it instead of how to implement the detailed casting. I have tried

template<template<typename T, int dimension> class Array, typename New_T, int dimension>

Array<typename New_T, int dimension> cast(Array<typename T, int dimension> a){



// detailed implementation of casting, which I do not care for this question.



}

but it cannot pass the compilation.

c++ templates template-deduction template-templates

edited Nov 20 '18 at 18:19

max66

35.8k74165

asked Jun 27 '18 at 2:05

Nan

577

edited Nov 20 '18 at 18:19

max66

35.8k74165

asked Jun 27 '18 at 2:05

Nan

577

edited Nov 20 '18 at 18:19

max66

35.8k74165

edited Nov 20 '18 at 18:19

max66

35.8k74165

edited Nov 20 '18 at 18:19

max66

35.8k74165

asked Jun 27 '18 at 2:05

Nan

577

asked Jun 27 '18 at 2:05

Nan

577

asked Jun 27 '18 at 2:05

Nan

577

Without having tested it for myself, I'm not sure you actually need a template template here. I think the function can just be: template<typename T, int dimension, typename New_T> Array<New_T, dimension> cast(Array<T, dimension>)

– John Ilacqua
Jun 27 '18 at 2:18

add a comment |

Without having tested it for myself, I'm not sure you actually need a template template here. I think the function can just be: template<typename T, int dimension, typename New_T> Array<New_T, dimension> cast(Array<T, dimension>)

– John Ilacqua
Jun 27 '18 at 2:18

Without having tested it for myself, I'm not sure you actually need a template template here. I think the function can just be: template<typename T, int dimension, typename New_T> Array<New_T, dimension> cast(Array<T, dimension>)

– John Ilacqua
Jun 27 '18 at 2:18

add a comment |

2 Answers
2

active

oldest

votes

How should I write this function? I am more interested in how to define it instead of how to implement the detailed casting.

I suppose something like

template <typename ToT, typename FromT, int Dim>

Array<ToT, Dim> cast (Array<FromT, Dim> const & inA)

 {

   // ...

 }

It's useful place ToT (to-type) in first position so you can explicit it and let FromT and Dim deduced from the inA value.

--- EDIT ---

The OP asks

Any insight why I have to put it [ToT] in the first position?

You don't necessarily have to put ToT in first position. But this simplify your life.

The point is that FromT and Dim are deducible from the inA argument; ToT isn't deducible from arguments so you have to explicit it.

But if you want to explicit a template parameter, you necessarily have to explicit the preceding parameters. So if you put ToT in last position, you have to call cast() explicating all template parameters

cast<int, 2, float>(a);

If you place ToT in first position, you have to explicit only it and leave the compiler deduce FromT and Dim from the argument

cast<float>(a);

edited Jun 27 '18 at 18:21

answered Jun 27 '18 at 2:33

max66

35.8k74165

Thanks for your suggestion. For me, the catch is that I have to put ToT (to-type) in the first position. I did that and it fixed my problem. Any insight why I have to put it in the fist position?

– Nan
Jun 27 '18 at 17:17

"fist position" :( ? Peace and love, Nan! (sorry: I'm joking; but I've modified my answer trying to respond to your last question).

– max66
Jun 27 '18 at 18:21

add a comment |

You don't need template template parameters at all here. Simple typename and int parameters will do:

template <typename T, int dimension>

class Array

{

    // ...

};



template <typename NewT, typename T, int dimension>

Array<NewT, dimension> cast(const Array<T, dimension>& a)

{

    // ...

}

Live Demo

You only need template template parameters when you want to accept different types of templates. For instance, if you wanted cast to be able to accept an Array or a std::array, you could use a template template parameter:

template<typename NewT, typename T, auto dimension, template<typename, auto> typename ArrayT>

ArrayT<NewT, dimension> cast(const ArrayT<T, dimension>& a)

{

    // ...

}

Live Demo

Note in this case I also changed the type of dimension to auto since std::array uses a size_t for its dimension while your Array uses int.

answered Jun 27 '18 at 2:34

Miles Budnek

10.6k11424

Nice explanation! I tried to use the template approach at first and it failed. I found out that the reason, as mentioned by @max66, was because I did not put the NewT type at the first position in the template.

– Nan
Jun 27 '18 at 17:19

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

How should I write this function? I am more interested in how to define it instead of how to implement the detailed casting.

I suppose something like

template <typename ToT, typename FromT, int Dim>

Array<ToT, Dim> cast (Array<FromT, Dim> const & inA)

 {

   // ...

 }

It's useful place ToT (to-type) in first position so you can explicit it and let FromT and Dim deduced from the inA value.

--- EDIT ---

The OP asks

Any insight why I have to put it [ToT] in the first position?

You don't necessarily have to put ToT in first position. But this simplify your life.

The point is that FromT and Dim are deducible from the inA argument; ToT isn't deducible from arguments so you have to explicit it.

cast<int, 2, float>(a);

If you place ToT in first position, you have to explicit only it and leave the compiler deduce FromT and Dim from the argument

cast<float>(a);

edited Jun 27 '18 at 18:21

answered Jun 27 '18 at 2:33

max66

35.8k74165

Thanks for your suggestion. For me, the catch is that I have to put ToT (to-type) in the first position. I did that and it fixed my problem. Any insight why I have to put it in the fist position?

– Nan
Jun 27 '18 at 17:17

"fist position" :( ? Peace and love, Nan! (sorry: I'm joking; but I've modified my answer trying to respond to your last question).

– max66
Jun 27 '18 at 18:21

add a comment |

How should I write this function? I am more interested in how to define it instead of how to implement the detailed casting.

I suppose something like

template <typename ToT, typename FromT, int Dim>

Array<ToT, Dim> cast (Array<FromT, Dim> const & inA)

 {

   // ...

 }

It's useful place ToT (to-type) in first position so you can explicit it and let FromT and Dim deduced from the inA value.

--- EDIT ---

The OP asks

Any insight why I have to put it [ToT] in the first position?

You don't necessarily have to put ToT in first position. But this simplify your life.

The point is that FromT and Dim are deducible from the inA argument; ToT isn't deducible from arguments so you have to explicit it.

cast<int, 2, float>(a);

If you place ToT in first position, you have to explicit only it and leave the compiler deduce FromT and Dim from the argument

cast<float>(a);

edited Jun 27 '18 at 18:21

answered Jun 27 '18 at 2:33

max66

35.8k74165

Thanks for your suggestion. For me, the catch is that I have to put ToT (to-type) in the first position. I did that and it fixed my problem. Any insight why I have to put it in the fist position?

– Nan
Jun 27 '18 at 17:17

"fist position" :( ? Peace and love, Nan! (sorry: I'm joking; but I've modified my answer trying to respond to your last question).

– max66
Jun 27 '18 at 18:21

add a comment |

How should I write this function? I am more interested in how to define it instead of how to implement the detailed casting.

I suppose something like

template <typename ToT, typename FromT, int Dim>

Array<ToT, Dim> cast (Array<FromT, Dim> const & inA)

 {

   // ...

 }

It's useful place ToT (to-type) in first position so you can explicit it and let FromT and Dim deduced from the inA value.

--- EDIT ---

The OP asks

Any insight why I have to put it [ToT] in the first position?

You don't necessarily have to put ToT in first position. But this simplify your life.

The point is that FromT and Dim are deducible from the inA argument; ToT isn't deducible from arguments so you have to explicit it.

cast<int, 2, float>(a);

If you place ToT in first position, you have to explicit only it and leave the compiler deduce FromT and Dim from the argument

cast<float>(a);

edited Jun 27 '18 at 18:21

answered Jun 27 '18 at 2:33

max66

35.8k74165

How should I write this function? I am more interested in how to define it instead of how to implement the detailed casting.

I suppose something like

template <typename ToT, typename FromT, int Dim>

Array<ToT, Dim> cast (Array<FromT, Dim> const & inA)

 {

   // ...

 }

It's useful place ToT (to-type) in first position so you can explicit it and let FromT and Dim deduced from the inA value.

--- EDIT ---

The OP asks

Any insight why I have to put it [ToT] in the first position?

You don't necessarily have to put ToT in first position. But this simplify your life.

The point is that FromT and Dim are deducible from the inA argument; ToT isn't deducible from arguments so you have to explicit it.

cast<int, 2, float>(a);

If you place ToT in first position, you have to explicit only it and leave the compiler deduce FromT and Dim from the argument

cast<float>(a);

edited Jun 27 '18 at 18:21

answered Jun 27 '18 at 2:33

max66

35.8k74165

edited Jun 27 '18 at 18:21

answered Jun 27 '18 at 2:33

max66

35.8k74165

answered Jun 27 '18 at 2:33

max66

35.8k74165

answered Jun 27 '18 at 2:33

max66

35.8k74165

Thanks for your suggestion. For me, the catch is that I have to put ToT (to-type) in the first position. I did that and it fixed my problem. Any insight why I have to put it in the fist position?

– Nan
Jun 27 '18 at 17:17

"fist position" :( ? Peace and love, Nan! (sorry: I'm joking; but I've modified my answer trying to respond to your last question).

– max66
Jun 27 '18 at 18:21

add a comment |

Thanks for your suggestion. For me, the catch is that I have to put ToT (to-type) in the first position. I did that and it fixed my problem. Any insight why I have to put it in the fist position?

– Nan
Jun 27 '18 at 17:17

"fist position" :( ? Peace and love, Nan! (sorry: I'm joking; but I've modified my answer trying to respond to your last question).

– max66
Jun 27 '18 at 18:21

Thanks for your suggestion. For me, the catch is that I have to put ToT (to-type) in the first position. I did that and it fixed my problem. Any insight why I have to put it in the fist position?

– Nan
Jun 27 '18 at 17:17

"fist position" :( ? Peace and love, Nan! (sorry: I'm joking; but I've modified my answer trying to respond to your last question).

– max66
Jun 27 '18 at 18:21

add a comment |

You don't need template template parameters at all here. Simple typename and int parameters will do:

template <typename T, int dimension>

class Array

{

    // ...

};



template <typename NewT, typename T, int dimension>

Array<NewT, dimension> cast(const Array<T, dimension>& a)

{

    // ...

}

Live Demo

template<typename NewT, typename T, auto dimension, template<typename, auto> typename ArrayT>

ArrayT<NewT, dimension> cast(const ArrayT<T, dimension>& a)

{

    // ...

}

Live Demo

Note in this case I also changed the type of dimension to auto since std::array uses a size_t for its dimension while your Array uses int.

answered Jun 27 '18 at 2:34

Miles Budnek

10.6k11424

Nice explanation! I tried to use the template approach at first and it failed. I found out that the reason, as mentioned by @max66, was because I did not put the NewT type at the first position in the template.

– Nan
Jun 27 '18 at 17:19

add a comment |

You don't need template template parameters at all here. Simple typename and int parameters will do:

template <typename T, int dimension>

class Array

{

    // ...

};



template <typename NewT, typename T, int dimension>

Array<NewT, dimension> cast(const Array<T, dimension>& a)

{

    // ...

}

Live Demo

template<typename NewT, typename T, auto dimension, template<typename, auto> typename ArrayT>

ArrayT<NewT, dimension> cast(const ArrayT<T, dimension>& a)

{

    // ...

}

Live Demo

Note in this case I also changed the type of dimension to auto since std::array uses a size_t for its dimension while your Array uses int.

answered Jun 27 '18 at 2:34

Miles Budnek

10.6k11424

Nice explanation! I tried to use the template approach at first and it failed. I found out that the reason, as mentioned by @max66, was because I did not put the NewT type at the first position in the template.

– Nan
Jun 27 '18 at 17:19

add a comment |

You don't need template template parameters at all here. Simple typename and int parameters will do:

template <typename T, int dimension>

class Array

{

    // ...

};



template <typename NewT, typename T, int dimension>

Array<NewT, dimension> cast(const Array<T, dimension>& a)

{

    // ...

}

Live Demo

template<typename NewT, typename T, auto dimension, template<typename, auto> typename ArrayT>

ArrayT<NewT, dimension> cast(const ArrayT<T, dimension>& a)

{

    // ...

}

Live Demo

Note in this case I also changed the type of dimension to auto since std::array uses a size_t for its dimension while your Array uses int.

answered Jun 27 '18 at 2:34

Miles Budnek

10.6k11424

You don't need template template parameters at all here. Simple typename and int parameters will do:

template <typename T, int dimension>

class Array

{

    // ...

};



template <typename NewT, typename T, int dimension>

Array<NewT, dimension> cast(const Array<T, dimension>& a)

{

    // ...

}

Live Demo

template<typename NewT, typename T, auto dimension, template<typename, auto> typename ArrayT>

ArrayT<NewT, dimension> cast(const ArrayT<T, dimension>& a)

{

    // ...

}

Live Demo

Note in this case I also changed the type of dimension to auto since std::array uses a size_t for its dimension while your Array uses int.

answered Jun 27 '18 at 2:34

Miles Budnek

10.6k11424

answered Jun 27 '18 at 2:34

Miles Budnek

10.6k11424

answered Jun 27 '18 at 2:34

Miles Budnek

10.6k11424

answered Jun 27 '18 at 2:34

Miles Budnek

10.6k11424

Nice explanation! I tried to use the template approach at first and it failed. I found out that the reason, as mentioned by @max66, was because I did not put the NewT type at the first position in the template.

– Nan
Jun 27 '18 at 17:19

add a comment |

Nice explanation! I tried to use the template approach at first and it failed. I found out that the reason, as mentioned by @max66, was because I did not put the NewT type at the first position in the template.

– Nan
Jun 27 '18 at 17:19

Nice explanation! I tried to use the template approach at first and it failed. I found out that the reason, as mentioned by @max66, was because I did not put the NewT type at the first position in the template.

– Nan
Jun 27 '18 at 17:19

add a comment |

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