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std::declval vs crtp, cannot deduce method return type from incomplete type
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std::declval vs crtp, cannot deduce method return type from incomplete type
Special behavior for decltype of call operator for incomplete typesC++11 does not deduce type when std::function or lambda functions are involvedgcc 4.7 about Variadic Templates/ decltype /std::forwardstd::declval() firing assertion error with warnings in GCCInferring return type of templated member functions in CRTPWhy doesn't std::shared_ptr need to know complete type if it's constructed from non-null?Specialize function template with decltype trailing return typeCan you declare a member variable with decltype on an object function?Invalid use of incomplete type struct std::hash with unordered_map with std::pair of enum class as keymixing CRTP with SFINAEno type named “type” in “std::result_of” ; get return type from overloading functions
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
I am trying to do something like this (in c++11):
#include <utility>
template <typename T>
struct base
using type = decltype( std::declval<T>().foo() );
;
struct bar : base<bar>
int foo() return 42;
;
int main()
bar::type x;
which fails with
prog.cc: In instantiation of 'struct base<bar>':
prog.cc:8:14: required from here
prog.cc:5:46: error: invalid use of incomplete type 'struct bar'
using type = decltype( std::declval<T>().foo() );
~~~~~~~~~~~~~~~~~~^~~
prog.cc:8:8: note: forward declaration of 'struct bar'
struct bar : base<bar> {
^~~
How can I declare an alias to the return type of bar::foo in base ? Is it not possible?
This question seems to be rather related: Special behavior for decltype of call operator for incomplete types, though I couldnt manage to apply the answer given there to my case.
c++ c++11 decltype crtp declval
asked 8 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
add a comment |
I am trying to do something like this (in c++11):
#include <utility>
template <typename T>
struct base
using type = decltype( std::declval<T>().foo() );
;
struct bar : base<bar>
int foo() return 42;
;
int main()
bar::type x;
which fails with
prog.cc: In instantiation of 'struct base<bar>':
prog.cc:8:14: required from here
prog.cc:5:46: error: invalid use of incomplete type 'struct bar'
using type = decltype( std::declval<T>().foo() );
~~~~~~~~~~~~~~~~~~^~~
prog.cc:8:8: note: forward declaration of 'struct bar'
struct bar : base<bar> {
^~~
How can I declare an alias to the return type of bar::foo in base ? Is it not possible?
This question seems to be rather related: Special behavior for decltype of call operator for incomplete types, though I couldnt manage to apply the answer given there to my case.
c++ c++11 decltype crtp declval
asked 8 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
add a comment |
I am trying to do something like this (in c++11):
#include <utility>
template <typename T>
struct base
using type = decltype( std::declval<T>().foo() );
;
struct bar : base<bar>
int foo() return 42;
;
int main()
bar::type x;
which fails with
prog.cc: In instantiation of 'struct base<bar>':
prog.cc:8:14: required from here
prog.cc:5:46: error: invalid use of incomplete type 'struct bar'
using type = decltype( std::declval<T>().foo() );
~~~~~~~~~~~~~~~~~~^~~
prog.cc:8:8: note: forward declaration of 'struct bar'
struct bar : base<bar> {
^~~
How can I declare an alias to the return type of bar::foo in base ? Is it not possible?
This question seems to be rather related: Special behavior for decltype of call operator for incomplete types, though I couldnt manage to apply the answer given there to my case.
c++ c++11 decltype crtp declval
asked 8 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
I am trying to do something like this (in c++11):
#include <utility>
template <typename T>
struct base
using type = decltype( std::declval<T>().foo() );
;
struct bar : base<bar>
int foo() return 42;
;
int main()
bar::type x;
which fails with
prog.cc: In instantiation of 'struct base<bar>':
prog.cc:8:14: required from here
prog.cc:5:46: error: invalid use of incomplete type 'struct bar'
using type = decltype( std::declval<T>().foo() );
~~~~~~~~~~~~~~~~~~^~~
prog.cc:8:8: note: forward declaration of 'struct bar'
struct bar : base<bar> {
^~~
How can I declare an alias to the return type of bar::foo in base ? Is it not possible?
This question seems to be rather related: Special behavior for decltype of call operator for incomplete types, though I couldnt manage to apply the answer given there to my case.
c++ c++11 decltype crtp declval
c++ c++11 decltype crtp declval
asked 8 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
asked 8 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
edited 5 hours ago
formerlyknownas_463035818
asked 8 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
asked 8 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
asked 8 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
21.4k43075
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You can make type a template type alias, so that users can instantiate it after the definition of bar is available. This will change the final syntax from bar::type to bar::type<>.
template <typename T>
struct base
template <typename G = T>
using type = decltype( std::declval<G>().foo() );
;
struct bar : base<bar>
int foo() return 42;
;
int main()
bar::type<> x;
live example on godbolt.org
answered 7 hours ago
Vittorio RomeoVittorio Romeo
61.4k17171318
I think now I also understand the reasoning in the answer I linked :) I dont like thebar::type<>too much but I could live with that
– formerlyknownas_463035818
7 hours ago
Unfortunately,type<>is not that useful inside a class scope. E.g., you can't writetype<> foo2();for abase's ofbar's member function without using similarG = Ttrick.
– Evg
7 hours ago
1
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
7 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can make type a template type alias, so that users can instantiate it after the definition of bar is available. This will change the final syntax from bar::type to bar::type<>.
template <typename T>
struct base
template <typename G = T>
using type = decltype( std::declval<G>().foo() );
;
struct bar : base<bar>
int foo() return 42;
;
int main()
bar::type<> x;
live example on godbolt.org
answered 7 hours ago
Vittorio RomeoVittorio Romeo
61.4k17171318
I think now I also understand the reasoning in the answer I linked :) I dont like thebar::type<>too much but I could live with that
– formerlyknownas_463035818
7 hours ago
Unfortunately,type<>is not that useful inside a class scope. E.g., you can't writetype<> foo2();for abase's ofbar's member function without using similarG = Ttrick.
– Evg
7 hours ago
1
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
7 hours ago
add a comment |
You can make type a template type alias, so that users can instantiate it after the definition of bar is available. This will change the final syntax from bar::type to bar::type<>.
template <typename T>
struct base
template <typename G = T>
using type = decltype( std::declval<G>().foo() );
;
struct bar : base<bar>
int foo() return 42;
;
int main()
bar::type<> x;
live example on godbolt.org
answered 7 hours ago
Vittorio RomeoVittorio Romeo
61.4k17171318
I think now I also understand the reasoning in the answer I linked :) I dont like thebar::type<>too much but I could live with that
– formerlyknownas_463035818
7 hours ago
Unfortunately,type<>is not that useful inside a class scope. E.g., you can't writetype<> foo2();for abase's ofbar's member function without using similarG = Ttrick.
– Evg
7 hours ago
1
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
7 hours ago
add a comment |
You can make type a template type alias, so that users can instantiate it after the definition of bar is available. This will change the final syntax from bar::type to bar::type<>.
template <typename T>
struct base
template <typename G = T>
using type = decltype( std::declval<G>().foo() );
;
struct bar : base<bar>
int foo() return 42;
;
int main()
bar::type<> x;
live example on godbolt.org
answered 7 hours ago
Vittorio RomeoVittorio Romeo
61.4k17171318
You can make type a template type alias, so that users can instantiate it after the definition of bar is available. This will change the final syntax from bar::type to bar::type<>.
template <typename T>
struct base
template <typename G = T>
using type = decltype( std::declval<G>().foo() );
;
struct bar : base<bar>
int foo() return 42;
;
int main()
bar::type<> x;
live example on godbolt.org
answered 7 hours ago
Vittorio RomeoVittorio Romeo
61.4k17171318
answered 7 hours ago
Vittorio RomeoVittorio Romeo
61.4k17171318
answered 7 hours ago
Vittorio RomeoVittorio Romeo
61.4k17171318
answered 7 hours ago
Vittorio RomeoVittorio Romeo
61.4k17171318
61.4k17171318
I think now I also understand the reasoning in the answer I linked :) I dont like thebar::type<>too much but I could live with that
– formerlyknownas_463035818
7 hours ago
Unfortunately,type<>is not that useful inside a class scope. E.g., you can't writetype<> foo2();for abase's ofbar's member function without using similarG = Ttrick.
– Evg
7 hours ago
1
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
7 hours ago
add a comment |
I think now I also understand the reasoning in the answer I linked :) I dont like thebar::type<>too much but I could live with that
– formerlyknownas_463035818
7 hours ago
Unfortunately,type<>is not that useful inside a class scope. E.g., you can't writetype<> foo2();for abase's ofbar's member function without using similarG = Ttrick.
– Evg
7 hours ago
1
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
7 hours ago
I think now I also understand the reasoning in the answer I linked :) I dont like the
bar::type<> too much but I could live with that– formerlyknownas_463035818
7 hours ago
I think now I also understand the reasoning in the answer I linked :) I dont like the
bar::type<> too much but I could live with that– formerlyknownas_463035818
7 hours ago
Unfortunately,
type<> is not that useful inside a class scope. E.g., you can't write type<> foo2(); for a base's of bar's member function without using similar G = T trick.– Evg
7 hours ago
Unfortunately,
type<> is not that useful inside a class scope. E.g., you can't write type<> foo2(); for a base's of bar's member function without using similar G = T trick.– Evg
7 hours ago
1
1
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
7 hours ago
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
7 hours ago
add a comment |
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