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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

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8

0

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

share|improve this question

edited 5 hours ago

formerlyknownas_463035818

asked 8 hours ago

formerlyknownas_463035818formerlyknownas_463035818

21.4k43075

add a comment | 

8

0

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

share|improve this question

edited 5 hours ago

formerlyknownas_463035818

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

share|improve this question

edited 5 hours ago

formerlyknownas_463035818

asked 8 hours ago

formerlyknownas_463035818formerlyknownas_463035818

21.4k43075

#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;

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> {
 ^~~

share|improve this question

edited 5 hours ago

formerlyknownas_463035818

asked 8 hours ago

formerlyknownas_463035818formerlyknownas_463035818

21.4k43075

share|improve this question

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

1 Answer
1

active

oldest

votes

10

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

share|improve this answer

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 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
  
  

  
  
  
  


• 
  
  
  

  
  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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10

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

share|improve this answer

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 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
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Evg: true, but you can still use it like this: gcc.godbolt.org/z/fHUxO5
  
  – Vittorio Romeo
  7 hours ago
  

  
  
  
  

add a comment | 

10

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

share|improve this answer

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 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
  
  

  
  
  
  


• 
  
  
  

  
  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

share|improve this answer

answered 7 hours ago

Vittorio RomeoVittorio Romeo

61.4k17171318

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;

share|improve this answer

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