generating functions at compile time

I am trying to generate functions at compile time using boost hana. Here is the code I wrote

#include <boost/hana/transform.hpp>

#include <array>

template<int i>
double f(double x)

 return x * i; 


int main()

 constexpr std::array arr = 1,5,10,100,500;

 constexpr auto functions = hana::transform(arr, 
 (const int a) -> double (*)(double)
 
 return f<a>;
 
 );

when compiling I get the error that f is not convertible to type double (*)(double).

I think the problem is that a is not constexpr(which is not possible since it is a function argument). Is there a way to make this working?

edited Nov 12 '18 at 21:18

asked Nov 12 '18 at 21:08

user3726947

304210

No, there is no way to instantiate a template with a runtime value.
– Galik
Nov 12 '18 at 21:18

but the array is known at compile time...
– user3726947
Nov 12 '18 at 21:22

Well it would be interesting if you wrote a constexpr function and used that instead of your lambda. That might work (I don't know how far constexpr is willing to introspect). Aslo is hana::transform constexpr? Probably not...
– Galik
Nov 12 '18 at 21:46

@galik - "That might work" - Have you tried? If the a value (the value used for f<a>) is passed through a template parameter, should works, constexpr or not constexpr. But if a is passed as normal parameter, shouldn't works, also inside a constexpr function, because a constexpr function can be called also run-time.
– max66
Nov 12 '18 at 22:48

Also Lambdas can be constexpr so this makes no diffrence
– user3726947
Nov 13 '18 at 23:35

add a comment |

I am trying to generate functions at compile time using boost hana. Here is the code I wrote

#include <boost/hana/transform.hpp>

#include <array>

template<int i>
double f(double x)

 return x * i; 


int main()

 constexpr std::array arr = 1,5,10,100,500;

 constexpr auto functions = hana::transform(arr, 
 (const int a) -> double (*)(double)
 
 return f<a>;
 
 );

when compiling I get the error that f is not convertible to type double (*)(double).

I think the problem is that a is not constexpr(which is not possible since it is a function argument). Is there a way to make this working?

edited Nov 12 '18 at 21:18

asked Nov 12 '18 at 21:08

user3726947

304210

No, there is no way to instantiate a template with a runtime value.
– Galik
Nov 12 '18 at 21:18

but the array is known at compile time...
– user3726947
Nov 12 '18 at 21:22

Well it would be interesting if you wrote a constexpr function and used that instead of your lambda. That might work (I don't know how far constexpr is willing to introspect). Aslo is hana::transform constexpr? Probably not...
– Galik
Nov 12 '18 at 21:46

@galik - "That might work" - Have you tried? If the a value (the value used for f<a>) is passed through a template parameter, should works, constexpr or not constexpr. But if a is passed as normal parameter, shouldn't works, also inside a constexpr function, because a constexpr function can be called also run-time.
– max66
Nov 12 '18 at 22:48

Also Lambdas can be constexpr so this makes no diffrence
– user3726947
Nov 13 '18 at 23:35

add a comment |

I am trying to generate functions at compile time using boost hana. Here is the code I wrote

#include <boost/hana/transform.hpp>

#include <array>

template<int i>
double f(double x)

 return x * i; 


int main()

 constexpr std::array arr = 1,5,10,100,500;

 constexpr auto functions = hana::transform(arr, 
 (const int a) -> double (*)(double)
 
 return f<a>;
 
 );

when compiling I get the error that f is not convertible to type double (*)(double).

I think the problem is that a is not constexpr(which is not possible since it is a function argument). Is there a way to make this working?

edited Nov 12 '18 at 21:18

asked Nov 12 '18 at 21:08

user3726947

304210

I am trying to generate functions at compile time using boost hana. Here is the code I wrote

#include <boost/hana/transform.hpp>

#include <array>

template<int i>
double f(double x)

 return x * i; 


int main()

 constexpr std::array arr = 1,5,10,100,500;

 constexpr auto functions = hana::transform(arr, 
 (const int a) -> double (*)(double)
 
 return f<a>;
 
 );

when compiling I get the error that f is not convertible to type double (*)(double).

I think the problem is that a is not constexpr(which is not possible since it is a function argument). Is there a way to make this working?

c++ templates metaprogramming

edited Nov 12 '18 at 21:18

asked Nov 12 '18 at 21:08

user3726947

304210

edited Nov 12 '18 at 21:18

asked Nov 12 '18 at 21:08

user3726947

304210

edited Nov 12 '18 at 21:18

asked Nov 12 '18 at 21:08

user3726947

304210

asked Nov 12 '18 at 21:08

user3726947

304210

asked Nov 12 '18 at 21:08

user3726947

304210

No, there is no way to instantiate a template with a runtime value.
– Galik
Nov 12 '18 at 21:18

but the array is known at compile time...
– user3726947
Nov 12 '18 at 21:22

Well it would be interesting if you wrote a constexpr function and used that instead of your lambda. That might work (I don't know how far constexpr is willing to introspect). Aslo is hana::transform constexpr? Probably not...
– Galik
Nov 12 '18 at 21:46

@galik - "That might work" - Have you tried? If the a value (the value used for f<a>) is passed through a template parameter, should works, constexpr or not constexpr. But if a is passed as normal parameter, shouldn't works, also inside a constexpr function, because a constexpr function can be called also run-time.
– max66
Nov 12 '18 at 22:48

Also Lambdas can be constexpr so this makes no diffrence
– user3726947
Nov 13 '18 at 23:35

add a comment |

No, there is no way to instantiate a template with a runtime value.
– Galik
Nov 12 '18 at 21:18

but the array is known at compile time...
– user3726947
Nov 12 '18 at 21:22

Well it would be interesting if you wrote a constexpr function and used that instead of your lambda. That might work (I don't know how far constexpr is willing to introspect). Aslo is hana::transform constexpr? Probably not...
– Galik
Nov 12 '18 at 21:46

@galik - "That might work" - Have you tried? If the a value (the value used for f<a>) is passed through a template parameter, should works, constexpr or not constexpr. But if a is passed as normal parameter, shouldn't works, also inside a constexpr function, because a constexpr function can be called also run-time.
– max66
Nov 12 '18 at 22:48

Also Lambdas can be constexpr so this makes no diffrence
– user3726947
Nov 13 '18 at 23:35

No, there is no way to instantiate a template with a runtime value.
– Galik
Nov 12 '18 at 21:18

but the array is known at compile time...
– user3726947
Nov 12 '18 at 21:22

Well it would be interesting if you wrote a constexpr function and used that instead of your lambda. That might work (I don't know how far constexpr is willing to introspect). Aslo is hana::transform constexpr? Probably not...
– Galik
Nov 12 '18 at 21:46

@galik - "That might work" - Have you tried? If the a value (the value used for f<a>) is passed through a template parameter, should works, constexpr or not constexpr. But if a is passed as normal parameter, shouldn't works, also inside a constexpr function, because a constexpr function can be called also run-time.
– max66
Nov 12 '18 at 22:48

Also Lambdas can be constexpr so this makes no diffrence
– user3726947
Nov 13 '18 at 23:35

add a comment |

1 Answer
1

active

oldest

votes

-1

Is there a way to make this working?

Not this way.

Look at your lambda

 (const int a) -> double (*)(double)
 
 return f<a>;

You're using the argument a, that is potentially known run-time in a lambda, as template parameter, that must be known compile type.

Can't work.

answered Nov 12 '18 at 21:19

max66

34.6k63762

so how can I make it work?
– user3726947
Nov 12 '18 at 21:20

1

@user3726947 - I don't see a way with std::array. Is good for you if you use std::integer_sequence; something as std::integer_sequence<int, 1, 5, 10, 100, 500>?
– max66
Nov 12 '18 at 21:22

ok i'll look into that, thanks!
– user3726947
Nov 12 '18 at 21:24

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

-1

Is there a way to make this working?

Not this way.

Look at your lambda

 (const int a) -> double (*)(double)
 
 return f<a>;

You're using the argument a, that is potentially known run-time in a lambda, as template parameter, that must be known compile type.

Can't work.

answered Nov 12 '18 at 21:19

max66

34.6k63762

so how can I make it work?
– user3726947
Nov 12 '18 at 21:20

1

@user3726947 - I don't see a way with std::array. Is good for you if you use std::integer_sequence; something as std::integer_sequence<int, 1, 5, 10, 100, 500>?
– max66
Nov 12 '18 at 21:22

ok i'll look into that, thanks!
– user3726947
Nov 12 '18 at 21:24

add a comment |

-1

Is there a way to make this working?

Not this way.

Look at your lambda

 (const int a) -> double (*)(double)
 
 return f<a>;

You're using the argument a, that is potentially known run-time in a lambda, as template parameter, that must be known compile type.

Can't work.

answered Nov 12 '18 at 21:19

max66

34.6k63762

so how can I make it work?
– user3726947
Nov 12 '18 at 21:20

1

@user3726947 - I don't see a way with std::array. Is good for you if you use std::integer_sequence; something as std::integer_sequence<int, 1, 5, 10, 100, 500>?
– max66
Nov 12 '18 at 21:22

ok i'll look into that, thanks!
– user3726947
Nov 12 '18 at 21:24

add a comment |

-1

Is there a way to make this working?

Not this way.

Look at your lambda

 (const int a) -> double (*)(double)
 
 return f<a>;

You're using the argument a, that is potentially known run-time in a lambda, as template parameter, that must be known compile type.

Can't work.

answered Nov 12 '18 at 21:19

max66

34.6k63762

Is there a way to make this working?

Not this way.

Look at your lambda

 (const int a) -> double (*)(double)
 
 return f<a>;

You're using the argument a, that is potentially known run-time in a lambda, as template parameter, that must be known compile type.

Can't work.

answered Nov 12 '18 at 21:19

max66

34.6k63762

answered Nov 12 '18 at 21:19

max66

34.6k63762

answered Nov 12 '18 at 21:19

max66

34.6k63762

answered Nov 12 '18 at 21:19

max66

34.6k63762

so how can I make it work?
– user3726947
Nov 12 '18 at 21:20

1

@user3726947 - I don't see a way with std::array. Is good for you if you use std::integer_sequence; something as std::integer_sequence<int, 1, 5, 10, 100, 500>?
– max66
Nov 12 '18 at 21:22

ok i'll look into that, thanks!
– user3726947
Nov 12 '18 at 21:24

add a comment |

so how can I make it work?
– user3726947
Nov 12 '18 at 21:20

1

@user3726947 - I don't see a way with std::array. Is good for you if you use std::integer_sequence; something as std::integer_sequence<int, 1, 5, 10, 100, 500>?
– max66
Nov 12 '18 at 21:22

ok i'll look into that, thanks!
– user3726947
Nov 12 '18 at 21:24

so how can I make it work?
– user3726947
Nov 12 '18 at 21:20

@user3726947 - I don't see a way with std::array. Is good for you if you use std::integer_sequence; something as std::integer_sequence<int, 1, 5, 10, 100, 500>?
– max66
Nov 12 '18 at 21:22

ok i'll look into that, thanks!
– user3726947
Nov 12 '18 at 21:24

add a comment |

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