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Is it possible to static_assert that a lambda is not generic?
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
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
add a comment |
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
1
What if the functor has an overloadedoperator()? 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
8 hours ago
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
7 hours ago
add a comment |
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
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
c++ c++17 variadic-templates template-meta-programming generic-lambda
edited 7 hours ago
max66
38.8k74473
38.8k74473
asked 9 hours ago
NoSenseEtAlNoSenseEtAl
7,7011677184
7,7011677184
1
What if the functor has an overloadedoperator()? 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
8 hours ago
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
7 hours ago
add a comment |
1
What if the functor has an overloadedoperator()? 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
8 hours ago
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
7 hours ago
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
8 hours ago
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
8 hours ago
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
7 hours ago
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
7 hours ago
add a comment |
3 Answers
3
active
oldest
votes
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.
1
Trying to take the address ofoperator()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
7 hours ago
add a comment |
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 :)
Nice idea ! You may ensure that "Auto" is unique like that :auto a_lambda = [](); using Auto= decltype(a_lambda);
– Martin m
5 hours ago
@Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?
– Konrad Rudolph
5 hours ago
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 m
5 hours ago
3
This gives both false positives (e.g.Visit([](std::any));) and false negatives (Visit([](int, auto));orVisit([](auto*));)
– Barry
3 hours ago
Visit([](int, auto))is not a valid case because ofv(std::get<Arg1>(data));andVisit([](auto*));is fixed now. I'm not sure ifVisit([](std::any))is one of the cases to be avoided as the question isn't clear enough.
– olivecoder
3 hours ago
|
show 3 more comments
#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.
add a comment |
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3 Answers
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3 Answers
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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.
1
Trying to take the address ofoperator()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
7 hours ago
add a comment |
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.
1
Trying to take the address ofoperator()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
7 hours ago
add a comment |
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.
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.
edited 3 hours ago
answered 7 hours ago
max66max66
38.8k74473
38.8k74473
1
Trying to take the address ofoperator()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
7 hours ago
add a comment |
1
Trying to take the address ofoperator()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
7 hours ago
1
1
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
7 hours ago
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
7 hours ago
add a comment |
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 :)
Nice idea ! You may ensure that "Auto" is unique like that :auto a_lambda = [](); using Auto= decltype(a_lambda);
– Martin m
5 hours ago
@Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?
– Konrad Rudolph
5 hours ago
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 m
5 hours ago
3
This gives both false positives (e.g.Visit([](std::any));) and false negatives (Visit([](int, auto));orVisit([](auto*));)
– Barry
3 hours ago
Visit([](int, auto))is not a valid case because ofv(std::get<Arg1>(data));andVisit([](auto*));is fixed now. I'm not sure ifVisit([](std::any))is one of the cases to be avoided as the question isn't clear enough.
– olivecoder
3 hours ago
|
show 3 more comments
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 :)
Nice idea ! You may ensure that "Auto" is unique like that :auto a_lambda = [](); using Auto= decltype(a_lambda);
– Martin m
5 hours ago
@Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?
– Konrad Rudolph
5 hours ago
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 m
5 hours ago
3
This gives both false positives (e.g.Visit([](std::any));) and false negatives (Visit([](int, auto));orVisit([](auto*));)
– Barry
3 hours ago
Visit([](int, auto))is not a valid case because ofv(std::get<Arg1>(data));andVisit([](auto*));is fixed now. I'm not sure ifVisit([](std::any))is one of the cases to be avoided as the question isn't clear enough.
– olivecoder
3 hours ago
|
show 3 more comments
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 :)
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 :)
edited 1 hour ago
answered 6 hours ago
olivecoderolivecoder
2,0681215
2,0681215
Nice idea ! You may ensure that "Auto" is unique like that :auto a_lambda = [](); using Auto= decltype(a_lambda);
– Martin m
5 hours ago
@Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?
– Konrad Rudolph
5 hours ago
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 m
5 hours ago
3
This gives both false positives (e.g.Visit([](std::any));) and false negatives (Visit([](int, auto));orVisit([](auto*));)
– Barry
3 hours ago
Visit([](int, auto))is not a valid case because ofv(std::get<Arg1>(data));andVisit([](auto*));is fixed now. I'm not sure ifVisit([](std::any))is one of the cases to be avoided as the question isn't clear enough.
– olivecoder
3 hours ago
|
show 3 more comments
Nice idea ! You may ensure that "Auto" is unique like that :auto a_lambda = [](); using Auto= decltype(a_lambda);
– Martin m
5 hours ago
@Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?
– Konrad Rudolph
5 hours ago
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 m
5 hours ago
3
This gives both false positives (e.g.Visit([](std::any));) and false negatives (Visit([](int, auto));orVisit([](auto*));)
– Barry
3 hours ago
Visit([](int, auto))is not a valid case because ofv(std::get<Arg1>(data));andVisit([](auto*));is fixed now. I'm not sure ifVisit([](std::any))is one of the cases to be avoided as the question isn't clear enough.
– olivecoder
3 hours ago
Nice idea ! You may ensure that "Auto" is unique like that :
auto a_lambda = [](); using Auto= decltype(a_lambda);– Martin m
5 hours ago
Nice idea ! You may ensure that "Auto" is unique like that :
auto a_lambda = [](); using Auto= decltype(a_lambda);– Martin m
5 hours ago
@Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?
– Konrad Rudolph
5 hours ago
@Martinm This seems to be the idiomatic way but does it actually have an advantage over the code in this answer?
– Konrad Rudolph
5 hours ago
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 m
5 hours ago
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 m
5 hours ago
3
3
This gives both false positives (e.g.
Visit([](std::any));) and false negatives (Visit([](int, auto)); or Visit([](auto*));)– Barry
3 hours ago
This gives both false positives (e.g.
Visit([](std::any));) and false negatives (Visit([](int, auto)); or Visit([](auto*));)– Barry
3 hours ago
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
3 hours ago
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
3 hours ago
|
show 3 more comments
#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.
add a comment |
#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.
add a comment |
#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.
#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.
answered 7 hours ago
Martin mMartin m
885
885
add a comment |
add a comment |
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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
8 hours ago
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
7 hours ago