std::declval vs crtp, cannot deduce method return type from incomplete typeSpecial behavior for decltype of...
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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
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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 10 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 10 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 10 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 10 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
asked 10 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
edited 7 hours ago
formerlyknownas_463035818
asked 10 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
asked 10 hours ago
formerlyknownas_463035818formerlyknownas_463035818
21.4k43075
asked 10 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 10 hours ago
Vittorio RomeoVittorio Romeo
61.5k17171318
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
10 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
10 hours ago
1
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
9 hours ago
add a comment |
Your Answer
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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 10 hours ago
Vittorio RomeoVittorio Romeo
61.5k17171318
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
10 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
10 hours ago
1
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
9 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 10 hours ago
Vittorio RomeoVittorio Romeo
61.5k17171318
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
10 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
10 hours ago
1
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
9 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 10 hours ago
Vittorio RomeoVittorio Romeo
61.5k17171318
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 10 hours ago
Vittorio RomeoVittorio Romeo
61.5k17171318
answered 10 hours ago
Vittorio RomeoVittorio Romeo
61.5k17171318
answered 10 hours ago
Vittorio RomeoVittorio Romeo
61.5k17171318
answered 10 hours ago
Vittorio RomeoVittorio Romeo
61.5k17171318
61.5k17171318
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
10 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
10 hours ago
1
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
9 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
10 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
10 hours ago
1
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
9 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
10 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
10 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
10 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
10 hours ago
1
1
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
9 hours ago
@Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
– Vittorio Romeo
9 hours ago
add a comment |
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