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Is it possible to static_assert that a lambda is not generic?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)
Data science time! April 2019 and salary with experience
The Ask Question Wizard is Live!Can we get the type of a lambda argument?Is it possible to write a template to check for a function's existence?What is a lambda expression in C++11?Compiling with gcc fails if using lambda function for QObject::connect()c++11 - getting result_of, decltype, std::function and variadic templates working togetherTemplate Type Deduction with LambdasVariadic template method and std::function - compilation errorRecursively visiting an `std::variant` using lambdas and fixed-point combinatorsClang can't find template binary operator in fold expressionGenerating lambda from class templateGetting active value in std::visit without knowing which value is active



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10















I implemented a Visit function (on a variant) that checks that the currently active type in the variant matches the function signature (more precisely the first argument). Based on this nice answer.
For example



#include <variant>
#include <string>
#include <iostream>

template<typename Ret, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(*) (Arg, Rest...));

template<typename Ret, typename F, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(F::*) (Arg, Rest...));

template<typename Ret, typename F, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(F::*) (Arg, Rest...) const);

template <typename F>
decltype(first_argument_helper(&F::operator())) first_argument_helper(F);

template <typename T>
using first_argument = decltype(first_argument_helper(std::declval<T>()));

std::variant<int, std::string> data="abc";
template <typename V>
void Visit(V v)
using Arg1 = typename std::remove_const_t<std::remove_reference_t<first_argument<V>>>;//... TMP magic to get 1st argument of visitor + remove cvr, see Q 43526647
if (! std::holds_alternative<Arg1>(data))
std::cerr<< "alternative mismatchn";
return;

v(std::get<Arg1>(data));

int main()
Visit([](const int& i)std::cout << i << "n"; );
Visit([](const std::string& s)std::cout << s << "n"; );
// Visit([](auto& x)); ugly kabooom



This works, but it explodes with a user unfriendly compile time error when users passes a generic (e.g. [](auto&)) lambda. Is there a way to detect this and give nice static_assert() about it?
Would also be nice if it worked with function templates as well, not just with lambdas.



Note that I do not know what possible lambdas do, so I can not do some clever stuff with Dummy types since lambdas may invoke arbitrary functions on types.
In other words I can not try to call lambda in 2 std::void_t tests on int and std::string and if it works assume it is generic because they might try to call .BlaLol() on int and string.










share|improve this question



















  • 1





    What if the functor has an overloaded operator()? Visiting is also very commonly performed with overloaded functors (see example 4 here), do those have to be forbidden (or have to work)?

    – Max Langhof
    Apr 3 at 7:35












  • I think that is too hard to handle, but if it can be done that would be nice... so it is optional, not required.

    – NoSenseEtAl
    Apr 3 at 8:16

















10















I implemented a Visit function (on a variant) that checks that the currently active type in the variant matches the function signature (more precisely the first argument). Based on this nice answer.
For example



#include <variant>
#include <string>
#include <iostream>

template<typename Ret, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(*) (Arg, Rest...));

template<typename Ret, typename F, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(F::*) (Arg, Rest...));

template<typename Ret, typename F, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(F::*) (Arg, Rest...) const);

template <typename F>
decltype(first_argument_helper(&F::operator())) first_argument_helper(F);

template <typename T>
using first_argument = decltype(first_argument_helper(std::declval<T>()));

std::variant<int, std::string> data="abc";
template <typename V>
void Visit(V v)
using Arg1 = typename std::remove_const_t<std::remove_reference_t<first_argument<V>>>;//... TMP magic to get 1st argument of visitor + remove cvr, see Q 43526647
if (! std::holds_alternative<Arg1>(data))
std::cerr<< "alternative mismatchn";
return;

v(std::get<Arg1>(data));

int main()
Visit([](const int& i)std::cout << i << "n"; );
Visit([](const std::string& s)std::cout << s << "n"; );
// Visit([](auto& x)); ugly kabooom



This works, but it explodes with a user unfriendly compile time error when users passes a generic (e.g. [](auto&)) lambda. Is there a way to detect this and give nice static_assert() about it?
Would also be nice if it worked with function templates as well, not just with lambdas.



Note that I do not know what possible lambdas do, so I can not do some clever stuff with Dummy types since lambdas may invoke arbitrary functions on types.
In other words I can not try to call lambda in 2 std::void_t tests on int and std::string and if it works assume it is generic because they might try to call .BlaLol() on int and string.










share|improve this question



















  • 1





    What if the functor has an overloaded operator()? Visiting is also very commonly performed with overloaded functors (see example 4 here), do those have to be forbidden (or have to work)?

    – Max Langhof
    Apr 3 at 7:35












  • I think that is too hard to handle, but if it can be done that would be nice... so it is optional, not required.

    – NoSenseEtAl
    Apr 3 at 8:16













10












10








10


2






I implemented a Visit function (on a variant) that checks that the currently active type in the variant matches the function signature (more precisely the first argument). Based on this nice answer.
For example



#include <variant>
#include <string>
#include <iostream>

template<typename Ret, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(*) (Arg, Rest...));

template<typename Ret, typename F, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(F::*) (Arg, Rest...));

template<typename Ret, typename F, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(F::*) (Arg, Rest...) const);

template <typename F>
decltype(first_argument_helper(&F::operator())) first_argument_helper(F);

template <typename T>
using first_argument = decltype(first_argument_helper(std::declval<T>()));

std::variant<int, std::string> data="abc";
template <typename V>
void Visit(V v)
using Arg1 = typename std::remove_const_t<std::remove_reference_t<first_argument<V>>>;//... TMP magic to get 1st argument of visitor + remove cvr, see Q 43526647
if (! std::holds_alternative<Arg1>(data))
std::cerr<< "alternative mismatchn";
return;

v(std::get<Arg1>(data));

int main()
Visit([](const int& i)std::cout << i << "n"; );
Visit([](const std::string& s)std::cout << s << "n"; );
// Visit([](auto& x)); ugly kabooom



This works, but it explodes with a user unfriendly compile time error when users passes a generic (e.g. [](auto&)) lambda. Is there a way to detect this and give nice static_assert() about it?
Would also be nice if it worked with function templates as well, not just with lambdas.



Note that I do not know what possible lambdas do, so I can not do some clever stuff with Dummy types since lambdas may invoke arbitrary functions on types.
In other words I can not try to call lambda in 2 std::void_t tests on int and std::string and if it works assume it is generic because they might try to call .BlaLol() on int and string.










share|improve this question
















I implemented a Visit function (on a variant) that checks that the currently active type in the variant matches the function signature (more precisely the first argument). Based on this nice answer.
For example



#include <variant>
#include <string>
#include <iostream>

template<typename Ret, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(*) (Arg, Rest...));

template<typename Ret, typename F, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(F::*) (Arg, Rest...));

template<typename Ret, typename F, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(F::*) (Arg, Rest...) const);

template <typename F>
decltype(first_argument_helper(&F::operator())) first_argument_helper(F);

template <typename T>
using first_argument = decltype(first_argument_helper(std::declval<T>()));

std::variant<int, std::string> data="abc";
template <typename V>
void Visit(V v)
using Arg1 = typename std::remove_const_t<std::remove_reference_t<first_argument<V>>>;//... TMP magic to get 1st argument of visitor + remove cvr, see Q 43526647
if (! std::holds_alternative<Arg1>(data))
std::cerr<< "alternative mismatchn";
return;

v(std::get<Arg1>(data));

int main()
Visit([](const int& i)std::cout << i << "n"; );
Visit([](const std::string& s)std::cout << s << "n"; );
// Visit([](auto& x)); ugly kabooom



This works, but it explodes with a user unfriendly compile time error when users passes a generic (e.g. [](auto&)) lambda. Is there a way to detect this and give nice static_assert() about it?
Would also be nice if it worked with function templates as well, not just with lambdas.



Note that I do not know what possible lambdas do, so I can not do some clever stuff with Dummy types since lambdas may invoke arbitrary functions on types.
In other words I can not try to call lambda in 2 std::void_t tests on int and std::string and if it works assume it is generic because they might try to call .BlaLol() on int and string.







c++ c++17 variadic-templates template-meta-programming generic-lambda






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 3 at 8:16









max66

39.6k74575




39.6k74575










asked Apr 3 at 6:21









NoSenseEtAlNoSenseEtAl

7,7741678185




7,7741678185







  • 1





    What if the functor has an overloaded operator()? Visiting is also very commonly performed with overloaded functors (see example 4 here), do those have to be forbidden (or have to work)?

    – Max Langhof
    Apr 3 at 7:35












  • I think that is too hard to handle, but if it can be done that would be nice... so it is optional, not required.

    – NoSenseEtAl
    Apr 3 at 8:16












  • 1





    What if the functor has an overloaded operator()? Visiting is also very commonly performed with overloaded functors (see example 4 here), do those have to be forbidden (or have to work)?

    – Max Langhof
    Apr 3 at 7:35












  • I think that is too hard to handle, but if it can be done that would be nice... so it is optional, not required.

    – NoSenseEtAl
    Apr 3 at 8:16







1




1





What if the functor has an overloaded operator()? Visiting is also very commonly performed with overloaded functors (see example 4 here), do those have to be forbidden (or have to work)?

– Max Langhof
Apr 3 at 7:35






What if the functor has an overloaded operator()? Visiting is also very commonly performed with overloaded functors (see example 4 here), do those have to be forbidden (or have to work)?

– Max Langhof
Apr 3 at 7:35














I think that is too hard to handle, but if it can be done that would be nice... so it is optional, not required.

– NoSenseEtAl
Apr 3 at 8:16





I think that is too hard to handle, but if it can be done that would be nice... so it is optional, not required.

– NoSenseEtAl
Apr 3 at 8:16












3 Answers
3






active

oldest

votes


















11















Is there a way to detect this and give nice static_assert about it?




I suppose you can use SFINAE over operator() type.



Follows an example



#include <type_traits>

template <typename T>
constexpr auto foo (T const &)
-> decltype( &T::operator(), bool )
return true;

constexpr bool foo (...)
return false;

int main()

auto l1 = [](int) return 0; ;
auto l2 = [](auto) return 0; ;

static_assert( foo(l1), "!" );
static_assert( ! foo(l2), "!" );



Instead of a bool, you can return std::true_type (from foo() first version) or std::false_type (from second version) if you want to use it through decltype().




Would also be nice if it worked with function templates as well, not just with lambdas.




I don't think it's possible in a so simple way: a lambda (also a generic lambda) is an object; a template function isn't an object but a set of objects. You can pass an object to a function, not a set of objects.



But the preceding solution should works also for classes/structs with operator()s: when there is a single, non template, operator(), you should get 1 from foo(); otherwise (no operator(), more than one operator(), template operator()), foo() should return 0.






share|improve this answer




















  • 2





    Trying to take the address of operator() was my initial idea as well (hence the comment) but then I somehow got lost in SFINAE on actually calling it. Anyway, here are some extra test cases involving overloaded functors: godbolt.org/z/M319jo

    – Max Langhof
    Apr 3 at 8:32



















4














Yet another simpler option:



#include <type_traits>
...
template <typename V>
void Visit(V v)
class Auto ;
static_assert(!std::is_invocable<V, Auto&>::value);
static_assert(!std::is_invocable<V, Auto*>::value);
...



The Auto class is just an invented type impossible to occur in the V parameters. If V accepts Auto as an argument it must be a generic.



I tested in coliru and I can confirm the solution covers these cases:



Visit([](auto x)); // nice static assert
Visit([](auto *x)); // nice static assert
Visit([](auto &x)); // nice static assert
Visit([](auto &&x)); // nice static assert


I'm not sure if that would cover all the possible lambdas that you don't know which are :)






share|improve this answer

























  • Nice idea ! You may ensure that "Auto" is unique like that : auto a_lambda = [](); using Auto= decltype(a_lambda);

    – Martin Morterol
    Apr 3 at 9:58












  • @Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?

    – Konrad Rudolph
    Apr 3 at 10:07











  • Well, I am not sure what happens in the case where an "Auto" class already define somewhere else. In meta-programming context I prefer to be sure that my type cannot be in conflict in some obscure case.

    – Martin Morterol
    Apr 3 at 10:19






  • 4





    This gives both false positives (e.g. Visit([](std::any));) and false negatives (Visit([](int, auto)); or Visit([](auto*));)

    – Barry
    Apr 3 at 11:47












  • Visit([](int, auto)) is not a valid case because of v(std::get<Arg1>(data)); and Visit([](auto*)); is fixed now. I'm not sure if Visit([](std::any)) is one of the cases to be avoided as the question isn't clear enough.

    – olivecoder
    Apr 3 at 12:09



















3














#include <variant>
#include <string>
#include <iostream>

template <class U, typename T = void>
struct can_be_checked : public std::false_type ;

template <typename U>
struct can_be_checked<U, std::enable_if_t< std::is_function<U>::value > > : public std::true_type;

template <typename U>
struct can_be_checked<U, std::void_t<decltype(&U::operator())>> : public std::true_type;


template<typename Ret, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(*) (Arg, Rest...));

template<typename Ret, typename F, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(F::*) (Arg, Rest...));

template<typename Ret, typename F, typename Arg, typename... Rest>
Arg first_argument_helper(Ret(F::*) (Arg, Rest...) const);

template <typename F>
decltype(first_argument_helper(&F::operator())) first_argument_helper(F);

template <typename T>
using first_argument = decltype(first_argument_helper(std::declval<T>()));

std::variant<int, std::string> data="abc";


template <typename V>
void Visit(V v)
if constexpr ( can_be_checked<std::remove_pointer_t<decltype(v)>>::value )

using Arg1 = typename std::remove_const_t<std::remove_reference_t<first_argument<V>>>;//... TMP magic to get 1st argument of visitor + remove cvr, see Q 43526647
if (! std::holds_alternative<Arg1>(data))

std::cerr<< "alternative mismatchn";
return;

v(std::get<Arg1>(data));

else

std::cout << "it's a template / auto lambda " << std::endl;





template <class T>
void foo(const T& t)

std::cout <<t << " foo n";


void fooi(const int& t)

std::cout <<t << " fooi " << std::endl;


int main()
Visit([](const int& i)std::cout << i << std::endl; );
Visit([](const std::string& s)std::cout << s << std::endl; );
Visit([](auto& x)std::cout <<x << std::endl;); // it's a template / auto lambda*/
Visit(foo<int>);

Visit<decltype(fooi)>(fooi);
Visit(fooi);


// Visit(foo); // => fail ugly



I don't know if it's you want, but you can, with that static_assert if an auto lambda is passed as parameter.



I think it's not possible to do the same for template function, but not sure.






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    3 Answers
    3






    active

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    11















    Is there a way to detect this and give nice static_assert about it?




    I suppose you can use SFINAE over operator() type.



    Follows an example



    #include <type_traits>

    template <typename T>
    constexpr auto foo (T const &)
    -> decltype( &T::operator(), bool )
    return true;

    constexpr bool foo (...)
    return false;

    int main()

    auto l1 = [](int) return 0; ;
    auto l2 = [](auto) return 0; ;

    static_assert( foo(l1), "!" );
    static_assert( ! foo(l2), "!" );



    Instead of a bool, you can return std::true_type (from foo() first version) or std::false_type (from second version) if you want to use it through decltype().




    Would also be nice if it worked with function templates as well, not just with lambdas.




    I don't think it's possible in a so simple way: a lambda (also a generic lambda) is an object; a template function isn't an object but a set of objects. You can pass an object to a function, not a set of objects.



    But the preceding solution should works also for classes/structs with operator()s: when there is a single, non template, operator(), you should get 1 from foo(); otherwise (no operator(), more than one operator(), template operator()), foo() should return 0.






    share|improve this answer




















    • 2





      Trying to take the address of operator() was my initial idea as well (hence the comment) but then I somehow got lost in SFINAE on actually calling it. Anyway, here are some extra test cases involving overloaded functors: godbolt.org/z/M319jo

      – Max Langhof
      Apr 3 at 8:32
















    11















    Is there a way to detect this and give nice static_assert about it?




    I suppose you can use SFINAE over operator() type.



    Follows an example



    #include <type_traits>

    template <typename T>
    constexpr auto foo (T const &)
    -> decltype( &T::operator(), bool )
    return true;

    constexpr bool foo (...)
    return false;

    int main()

    auto l1 = [](int) return 0; ;
    auto l2 = [](auto) return 0; ;

    static_assert( foo(l1), "!" );
    static_assert( ! foo(l2), "!" );



    Instead of a bool, you can return std::true_type (from foo() first version) or std::false_type (from second version) if you want to use it through decltype().




    Would also be nice if it worked with function templates as well, not just with lambdas.




    I don't think it's possible in a so simple way: a lambda (also a generic lambda) is an object; a template function isn't an object but a set of objects. You can pass an object to a function, not a set of objects.



    But the preceding solution should works also for classes/structs with operator()s: when there is a single, non template, operator(), you should get 1 from foo(); otherwise (no operator(), more than one operator(), template operator()), foo() should return 0.






    share|improve this answer




















    • 2





      Trying to take the address of operator() was my initial idea as well (hence the comment) but then I somehow got lost in SFINAE on actually calling it. Anyway, here are some extra test cases involving overloaded functors: godbolt.org/z/M319jo

      – Max Langhof
      Apr 3 at 8:32














    11












    11








    11








    Is there a way to detect this and give nice static_assert about it?




    I suppose you can use SFINAE over operator() type.



    Follows an example



    #include <type_traits>

    template <typename T>
    constexpr auto foo (T const &)
    -> decltype( &T::operator(), bool )
    return true;

    constexpr bool foo (...)
    return false;

    int main()

    auto l1 = [](int) return 0; ;
    auto l2 = [](auto) return 0; ;

    static_assert( foo(l1), "!" );
    static_assert( ! foo(l2), "!" );



    Instead of a bool, you can return std::true_type (from foo() first version) or std::false_type (from second version) if you want to use it through decltype().




    Would also be nice if it worked with function templates as well, not just with lambdas.




    I don't think it's possible in a so simple way: a lambda (also a generic lambda) is an object; a template function isn't an object but a set of objects. You can pass an object to a function, not a set of objects.



    But the preceding solution should works also for classes/structs with operator()s: when there is a single, non template, operator(), you should get 1 from foo(); otherwise (no operator(), more than one operator(), template operator()), foo() should return 0.






    share|improve this answer
















    Is there a way to detect this and give nice static_assert about it?




    I suppose you can use SFINAE over operator() type.



    Follows an example



    #include <type_traits>

    template <typename T>
    constexpr auto foo (T const &)
    -> decltype( &T::operator(), bool )
    return true;

    constexpr bool foo (...)
    return false;

    int main()

    auto l1 = [](int) return 0; ;
    auto l2 = [](auto) return 0; ;

    static_assert( foo(l1), "!" );
    static_assert( ! foo(l2), "!" );



    Instead of a bool, you can return std::true_type (from foo() first version) or std::false_type (from second version) if you want to use it through decltype().




    Would also be nice if it worked with function templates as well, not just with lambdas.




    I don't think it's possible in a so simple way: a lambda (also a generic lambda) is an object; a template function isn't an object but a set of objects. You can pass an object to a function, not a set of objects.



    But the preceding solution should works also for classes/structs with operator()s: when there is a single, non template, operator(), you should get 1 from foo(); otherwise (no operator(), more than one operator(), template operator()), foo() should return 0.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Apr 3 at 11:43

























    answered Apr 3 at 8:06









    max66max66

    39.6k74575




    39.6k74575







    • 2





      Trying to take the address of operator() was my initial idea as well (hence the comment) but then I somehow got lost in SFINAE on actually calling it. Anyway, here are some extra test cases involving overloaded functors: godbolt.org/z/M319jo

      – Max Langhof
      Apr 3 at 8:32













    • 2





      Trying to take the address of operator() was my initial idea as well (hence the comment) but then I somehow got lost in SFINAE on actually calling it. Anyway, here are some extra test cases involving overloaded functors: godbolt.org/z/M319jo

      – Max Langhof
      Apr 3 at 8:32








    2




    2





    Trying to take the address of operator() was my initial idea as well (hence the comment) but then I somehow got lost in SFINAE on actually calling it. Anyway, here are some extra test cases involving overloaded functors: godbolt.org/z/M319jo

    – Max Langhof
    Apr 3 at 8:32






    Trying to take the address of operator() was my initial idea as well (hence the comment) but then I somehow got lost in SFINAE on actually calling it. Anyway, here are some extra test cases involving overloaded functors: godbolt.org/z/M319jo

    – Max Langhof
    Apr 3 at 8:32














    4














    Yet another simpler option:



    #include <type_traits>
    ...
    template <typename V>
    void Visit(V v)
    class Auto ;
    static_assert(!std::is_invocable<V, Auto&>::value);
    static_assert(!std::is_invocable<V, Auto*>::value);
    ...



    The Auto class is just an invented type impossible to occur in the V parameters. If V accepts Auto as an argument it must be a generic.



    I tested in coliru and I can confirm the solution covers these cases:



    Visit([](auto x)); // nice static assert
    Visit([](auto *x)); // nice static assert
    Visit([](auto &x)); // nice static assert
    Visit([](auto &&x)); // nice static assert


    I'm not sure if that would cover all the possible lambdas that you don't know which are :)






    share|improve this answer

























    • Nice idea ! You may ensure that "Auto" is unique like that : auto a_lambda = [](); using Auto= decltype(a_lambda);

      – Martin Morterol
      Apr 3 at 9:58












    • @Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?

      – Konrad Rudolph
      Apr 3 at 10:07











    • Well, I am not sure what happens in the case where an "Auto" class already define somewhere else. In meta-programming context I prefer to be sure that my type cannot be in conflict in some obscure case.

      – Martin Morterol
      Apr 3 at 10:19






    • 4





      This gives both false positives (e.g. Visit([](std::any));) and false negatives (Visit([](int, auto)); or Visit([](auto*));)

      – Barry
      Apr 3 at 11:47












    • Visit([](int, auto)) is not a valid case because of v(std::get<Arg1>(data)); and Visit([](auto*)); is fixed now. I'm not sure if Visit([](std::any)) is one of the cases to be avoided as the question isn't clear enough.

      – olivecoder
      Apr 3 at 12:09
















    4














    Yet another simpler option:



    #include <type_traits>
    ...
    template <typename V>
    void Visit(V v)
    class Auto ;
    static_assert(!std::is_invocable<V, Auto&>::value);
    static_assert(!std::is_invocable<V, Auto*>::value);
    ...



    The Auto class is just an invented type impossible to occur in the V parameters. If V accepts Auto as an argument it must be a generic.



    I tested in coliru and I can confirm the solution covers these cases:



    Visit([](auto x)); // nice static assert
    Visit([](auto *x)); // nice static assert
    Visit([](auto &x)); // nice static assert
    Visit([](auto &&x)); // nice static assert


    I'm not sure if that would cover all the possible lambdas that you don't know which are :)






    share|improve this answer

























    • Nice idea ! You may ensure that "Auto" is unique like that : auto a_lambda = [](); using Auto= decltype(a_lambda);

      – Martin Morterol
      Apr 3 at 9:58












    • @Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?

      – Konrad Rudolph
      Apr 3 at 10:07











    • Well, I am not sure what happens in the case where an "Auto" class already define somewhere else. In meta-programming context I prefer to be sure that my type cannot be in conflict in some obscure case.

      – Martin Morterol
      Apr 3 at 10:19






    • 4





      This gives both false positives (e.g. Visit([](std::any));) and false negatives (Visit([](int, auto)); or Visit([](auto*));)

      – Barry
      Apr 3 at 11:47












    • Visit([](int, auto)) is not a valid case because of v(std::get<Arg1>(data)); and Visit([](auto*)); is fixed now. I'm not sure if Visit([](std::any)) is one of the cases to be avoided as the question isn't clear enough.

      – olivecoder
      Apr 3 at 12:09














    4












    4








    4







    Yet another simpler option:



    #include <type_traits>
    ...
    template <typename V>
    void Visit(V v)
    class Auto ;
    static_assert(!std::is_invocable<V, Auto&>::value);
    static_assert(!std::is_invocable<V, Auto*>::value);
    ...



    The Auto class is just an invented type impossible to occur in the V parameters. If V accepts Auto as an argument it must be a generic.



    I tested in coliru and I can confirm the solution covers these cases:



    Visit([](auto x)); // nice static assert
    Visit([](auto *x)); // nice static assert
    Visit([](auto &x)); // nice static assert
    Visit([](auto &&x)); // nice static assert


    I'm not sure if that would cover all the possible lambdas that you don't know which are :)






    share|improve this answer















    Yet another simpler option:



    #include <type_traits>
    ...
    template <typename V>
    void Visit(V v)
    class Auto ;
    static_assert(!std::is_invocable<V, Auto&>::value);
    static_assert(!std::is_invocable<V, Auto*>::value);
    ...



    The Auto class is just an invented type impossible to occur in the V parameters. If V accepts Auto as an argument it must be a generic.



    I tested in coliru and I can confirm the solution covers these cases:



    Visit([](auto x)); // nice static assert
    Visit([](auto *x)); // nice static assert
    Visit([](auto &x)); // nice static assert
    Visit([](auto &&x)); // nice static assert


    I'm not sure if that would cover all the possible lambdas that you don't know which are :)







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Apr 3 at 14:05

























    answered Apr 3 at 8:56









    olivecoderolivecoder

    2,0881217




    2,0881217












    • Nice idea ! You may ensure that "Auto" is unique like that : auto a_lambda = [](); using Auto= decltype(a_lambda);

      – Martin Morterol
      Apr 3 at 9:58












    • @Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?

      – Konrad Rudolph
      Apr 3 at 10:07











    • Well, I am not sure what happens in the case where an "Auto" class already define somewhere else. In meta-programming context I prefer to be sure that my type cannot be in conflict in some obscure case.

      – Martin Morterol
      Apr 3 at 10:19






    • 4





      This gives both false positives (e.g. Visit([](std::any));) and false negatives (Visit([](int, auto)); or Visit([](auto*));)

      – Barry
      Apr 3 at 11:47












    • Visit([](int, auto)) is not a valid case because of v(std::get<Arg1>(data)); and Visit([](auto*)); is fixed now. I'm not sure if Visit([](std::any)) is one of the cases to be avoided as the question isn't clear enough.

      – olivecoder
      Apr 3 at 12:09


















    • Nice idea ! You may ensure that "Auto" is unique like that : auto a_lambda = [](); using Auto= decltype(a_lambda);

      – Martin Morterol
      Apr 3 at 9:58












    • @Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?

      – Konrad Rudolph
      Apr 3 at 10:07











    • Well, I am not sure what happens in the case where an "Auto" class already define somewhere else. In meta-programming context I prefer to be sure that my type cannot be in conflict in some obscure case.

      – Martin Morterol
      Apr 3 at 10:19






    • 4





      This gives both false positives (e.g. Visit([](std::any));) and false negatives (Visit([](int, auto)); or Visit([](auto*));)

      – Barry
      Apr 3 at 11:47












    • Visit([](int, auto)) is not a valid case because of v(std::get<Arg1>(data)); and Visit([](auto*)); is fixed now. I'm not sure if Visit([](std::any)) is one of the cases to be avoided as the question isn't clear enough.

      – olivecoder
      Apr 3 at 12:09

















    Nice idea ! You may ensure that "Auto" is unique like that : auto a_lambda = [](); using Auto= decltype(a_lambda);

    – Martin Morterol
    Apr 3 at 9:58






    Nice idea ! You may ensure that "Auto" is unique like that : auto a_lambda = [](); using Auto= decltype(a_lambda);

    – Martin Morterol
    Apr 3 at 9:58














    @Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?

    – Konrad Rudolph
    Apr 3 at 10:07





    @Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?

    – Konrad Rudolph
    Apr 3 at 10:07













    Well, I am not sure what happens in the case where an "Auto" class already define somewhere else. In meta-programming context I prefer to be sure that my type cannot be in conflict in some obscure case.

    – Martin Morterol
    Apr 3 at 10:19





    Well, I am not sure what happens in the case where an "Auto" class already define somewhere else. In meta-programming context I prefer to be sure that my type cannot be in conflict in some obscure case.

    – Martin Morterol
    Apr 3 at 10:19




    4




    4





    This gives both false positives (e.g. Visit([](std::any));) and false negatives (Visit([](int, auto)); or Visit([](auto*));)

    – Barry
    Apr 3 at 11:47






    This gives both false positives (e.g. Visit([](std::any));) and false negatives (Visit([](int, auto)); or Visit([](auto*));)

    – Barry
    Apr 3 at 11:47














    Visit([](int, auto)) is not a valid case because of v(std::get<Arg1>(data)); and Visit([](auto*)); is fixed now. I'm not sure if Visit([](std::any)) is one of the cases to be avoided as the question isn't clear enough.

    – olivecoder
    Apr 3 at 12:09






    Visit([](int, auto)) is not a valid case because of v(std::get<Arg1>(data)); and Visit([](auto*)); is fixed now. I'm not sure if Visit([](std::any)) is one of the cases to be avoided as the question isn't clear enough.

    – olivecoder
    Apr 3 at 12:09












    3














    #include <variant>
    #include <string>
    #include <iostream>

    template <class U, typename T = void>
    struct can_be_checked : public std::false_type ;

    template <typename U>
    struct can_be_checked<U, std::enable_if_t< std::is_function<U>::value > > : public std::true_type;

    template <typename U>
    struct can_be_checked<U, std::void_t<decltype(&U::operator())>> : public std::true_type;


    template<typename Ret, typename Arg, typename... Rest>
    Arg first_argument_helper(Ret(*) (Arg, Rest...));

    template<typename Ret, typename F, typename Arg, typename... Rest>
    Arg first_argument_helper(Ret(F::*) (Arg, Rest...));

    template<typename Ret, typename F, typename Arg, typename... Rest>
    Arg first_argument_helper(Ret(F::*) (Arg, Rest...) const);

    template <typename F>
    decltype(first_argument_helper(&F::operator())) first_argument_helper(F);

    template <typename T>
    using first_argument = decltype(first_argument_helper(std::declval<T>()));

    std::variant<int, std::string> data="abc";


    template <typename V>
    void Visit(V v)
    if constexpr ( can_be_checked<std::remove_pointer_t<decltype(v)>>::value )

    using Arg1 = typename std::remove_const_t<std::remove_reference_t<first_argument<V>>>;//... TMP magic to get 1st argument of visitor + remove cvr, see Q 43526647
    if (! std::holds_alternative<Arg1>(data))

    std::cerr<< "alternative mismatchn";
    return;

    v(std::get<Arg1>(data));

    else

    std::cout << "it's a template / auto lambda " << std::endl;





    template <class T>
    void foo(const T& t)

    std::cout <<t << " foo n";


    void fooi(const int& t)

    std::cout <<t << " fooi " << std::endl;


    int main()
    Visit([](const int& i)std::cout << i << std::endl; );
    Visit([](const std::string& s)std::cout << s << std::endl; );
    Visit([](auto& x)std::cout <<x << std::endl;); // it's a template / auto lambda*/
    Visit(foo<int>);

    Visit<decltype(fooi)>(fooi);
    Visit(fooi);


    // Visit(foo); // => fail ugly



    I don't know if it's you want, but you can, with that static_assert if an auto lambda is passed as parameter.



    I think it's not possible to do the same for template function, but not sure.






    share|improve this answer



























      3














      #include <variant>
      #include <string>
      #include <iostream>

      template <class U, typename T = void>
      struct can_be_checked : public std::false_type ;

      template <typename U>
      struct can_be_checked<U, std::enable_if_t< std::is_function<U>::value > > : public std::true_type;

      template <typename U>
      struct can_be_checked<U, std::void_t<decltype(&U::operator())>> : public std::true_type;


      template<typename Ret, typename Arg, typename... Rest>
      Arg first_argument_helper(Ret(*) (Arg, Rest...));

      template<typename Ret, typename F, typename Arg, typename... Rest>
      Arg first_argument_helper(Ret(F::*) (Arg, Rest...));

      template<typename Ret, typename F, typename Arg, typename... Rest>
      Arg first_argument_helper(Ret(F::*) (Arg, Rest...) const);

      template <typename F>
      decltype(first_argument_helper(&F::operator())) first_argument_helper(F);

      template <typename T>
      using first_argument = decltype(first_argument_helper(std::declval<T>()));

      std::variant<int, std::string> data="abc";


      template <typename V>
      void Visit(V v)
      if constexpr ( can_be_checked<std::remove_pointer_t<decltype(v)>>::value )

      using Arg1 = typename std::remove_const_t<std::remove_reference_t<first_argument<V>>>;//... TMP magic to get 1st argument of visitor + remove cvr, see Q 43526647
      if (! std::holds_alternative<Arg1>(data))

      std::cerr<< "alternative mismatchn";
      return;

      v(std::get<Arg1>(data));

      else

      std::cout << "it's a template / auto lambda " << std::endl;





      template <class T>
      void foo(const T& t)

      std::cout <<t << " foo n";


      void fooi(const int& t)

      std::cout <<t << " fooi " << std::endl;


      int main()
      Visit([](const int& i)std::cout << i << std::endl; );
      Visit([](const std::string& s)std::cout << s << std::endl; );
      Visit([](auto& x)std::cout <<x << std::endl;); // it's a template / auto lambda*/
      Visit(foo<int>);

      Visit<decltype(fooi)>(fooi);
      Visit(fooi);


      // Visit(foo); // => fail ugly



      I don't know if it's you want, but you can, with that static_assert if an auto lambda is passed as parameter.



      I think it's not possible to do the same for template function, but not sure.






      share|improve this answer

























        3












        3








        3







        #include <variant>
        #include <string>
        #include <iostream>

        template <class U, typename T = void>
        struct can_be_checked : public std::false_type ;

        template <typename U>
        struct can_be_checked<U, std::enable_if_t< std::is_function<U>::value > > : public std::true_type;

        template <typename U>
        struct can_be_checked<U, std::void_t<decltype(&U::operator())>> : public std::true_type;


        template<typename Ret, typename Arg, typename... Rest>
        Arg first_argument_helper(Ret(*) (Arg, Rest...));

        template<typename Ret, typename F, typename Arg, typename... Rest>
        Arg first_argument_helper(Ret(F::*) (Arg, Rest...));

        template<typename Ret, typename F, typename Arg, typename... Rest>
        Arg first_argument_helper(Ret(F::*) (Arg, Rest...) const);

        template <typename F>
        decltype(first_argument_helper(&F::operator())) first_argument_helper(F);

        template <typename T>
        using first_argument = decltype(first_argument_helper(std::declval<T>()));

        std::variant<int, std::string> data="abc";


        template <typename V>
        void Visit(V v)
        if constexpr ( can_be_checked<std::remove_pointer_t<decltype(v)>>::value )

        using Arg1 = typename std::remove_const_t<std::remove_reference_t<first_argument<V>>>;//... TMP magic to get 1st argument of visitor + remove cvr, see Q 43526647
        if (! std::holds_alternative<Arg1>(data))

        std::cerr<< "alternative mismatchn";
        return;

        v(std::get<Arg1>(data));

        else

        std::cout << "it's a template / auto lambda " << std::endl;





        template <class T>
        void foo(const T& t)

        std::cout <<t << " foo n";


        void fooi(const int& t)

        std::cout <<t << " fooi " << std::endl;


        int main()
        Visit([](const int& i)std::cout << i << std::endl; );
        Visit([](const std::string& s)std::cout << s << std::endl; );
        Visit([](auto& x)std::cout <<x << std::endl;); // it's a template / auto lambda*/
        Visit(foo<int>);

        Visit<decltype(fooi)>(fooi);
        Visit(fooi);


        // Visit(foo); // => fail ugly



        I don't know if it's you want, but you can, with that static_assert if an auto lambda is passed as parameter.



        I think it's not possible to do the same for template function, but not sure.






        share|improve this answer













        #include <variant>
        #include <string>
        #include <iostream>

        template <class U, typename T = void>
        struct can_be_checked : public std::false_type ;

        template <typename U>
        struct can_be_checked<U, std::enable_if_t< std::is_function<U>::value > > : public std::true_type;

        template <typename U>
        struct can_be_checked<U, std::void_t<decltype(&U::operator())>> : public std::true_type;


        template<typename Ret, typename Arg, typename... Rest>
        Arg first_argument_helper(Ret(*) (Arg, Rest...));

        template<typename Ret, typename F, typename Arg, typename... Rest>
        Arg first_argument_helper(Ret(F::*) (Arg, Rest...));

        template<typename Ret, typename F, typename Arg, typename... Rest>
        Arg first_argument_helper(Ret(F::*) (Arg, Rest...) const);

        template <typename F>
        decltype(first_argument_helper(&F::operator())) first_argument_helper(F);

        template <typename T>
        using first_argument = decltype(first_argument_helper(std::declval<T>()));

        std::variant<int, std::string> data="abc";


        template <typename V>
        void Visit(V v)
        if constexpr ( can_be_checked<std::remove_pointer_t<decltype(v)>>::value )

        using Arg1 = typename std::remove_const_t<std::remove_reference_t<first_argument<V>>>;//... TMP magic to get 1st argument of visitor + remove cvr, see Q 43526647
        if (! std::holds_alternative<Arg1>(data))

        std::cerr<< "alternative mismatchn";
        return;

        v(std::get<Arg1>(data));

        else

        std::cout << "it's a template / auto lambda " << std::endl;





        template <class T>
        void foo(const T& t)

        std::cout <<t << " foo n";


        void fooi(const int& t)

        std::cout <<t << " fooi " << std::endl;


        int main()
        Visit([](const int& i)std::cout << i << std::endl; );
        Visit([](const std::string& s)std::cout << s << std::endl; );
        Visit([](auto& x)std::cout <<x << std::endl;); // it's a template / auto lambda*/
        Visit(foo<int>);

        Visit<decltype(fooi)>(fooi);
        Visit(fooi);


        // Visit(foo); // => fail ugly



        I don't know if it's you want, but you can, with that static_assert if an auto lambda is passed as parameter.



        I think it's not possible to do the same for template function, but not sure.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Apr 3 at 8:12









        Martin MorterolMartin Morterol

        31316




        31316



























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