generating functions at compile time










0














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?










share|improve this question























  • 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
















0














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?










share|improve this question























  • 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














0












0








0


1





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?










share|improve this question















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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 '18 at 21:18

























asked Nov 12 '18 at 21:08









user3726947

304210




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

















  • 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





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




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





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




@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





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













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.






share|improve this answer




















  • 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










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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.






share|improve this answer




















  • 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















-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.






share|improve this answer




















  • 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













-1












-1








-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.






share|improve this answer













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.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 12 '18 at 21:19









max66

34.6k63762




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
















  • 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





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





1




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




@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




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

















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