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C++'s most vexing parse again [duplicate]

A confusing detail about the Most Vexing ParseC++ Why is variable a function and not an object?why c++ class initialisation from another class type not throwing an error?Calling a default constructor in a constructor argumentInitialization of vector with stream iteratorsCopy and Base Constructor don't initialize C++Prototyped class function not calledWhy the template parameter pack opens when passed Objects rather than typesConversion constructor not called where it should berequest for member “print” in “yello” which is of non-class type 'knight'What are the differences between a pointer variable and a reference variable in C++?How do I use arrays in C++?Pretty-print C++ STL containersHow can I convert a std::string to int?Does the GotW #101 “solution” actually solve anything?Image Processing: Algorithm Improvement for 'Coca-Cola Can' RecognitionIs there a way to write make_unique() in VS2012?Why initialize unique_ptr with a make_unique call?Replacing a 32-bit loop counter with 64-bit introduces crazy performance deviationsIs there a c++ temporary rvalue for constants

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39

12

This question already has an answer here:

• 
  A confusing detail about the Most Vexing Parse
  
  4 answers
  
  

Taken directly from http://herbsutter.com/2013/05/09/gotw-1-solution/

While widget w(); is clear for me, I have no idea how can the below code be a function declaration?

// same problem (gadget and doodad are types)
//
widget w( gadget(), doodad() ); // pitfall: not a variable declaration

How is this possible?

c++ c++11 gotw

share|improve this question

edited May 16 '13 at 15:21

0x499602D2

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asked May 16 '13 at 13:27

YankoYanko

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marked as duplicate by Nicol Bolas c++
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Jun 16 at 18:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

39

12

This question already has an answer here:

• 
  A confusing detail about the Most Vexing Parse
  
  4 answers
  
  

Taken directly from http://herbsutter.com/2013/05/09/gotw-1-solution/

While widget w(); is clear for me, I have no idea how can the below code be a function declaration?

// same problem (gadget and doodad are types)
//
widget w( gadget(), doodad() ); // pitfall: not a variable declaration

How is this possible?

c++ c++11 gotw

share|improve this question

edited May 16 '13 at 15:21

0x499602D2

69.7k2929 gold badges123123 silver badges209209 bronze badges

asked May 16 '13 at 13:27

YankoYanko

42455 silver badges1515 bronze badges

marked as duplicate by Nicol Bolas c++
Users with the  c++ badge can single-handedly close c++ questions as duplicates and reopen them as needed.

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Jun 16 at 18:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

This question already has an answer here:

• 
  A confusing detail about the Most Vexing Parse
  
  4 answers
  
  

Taken directly from http://herbsutter.com/2013/05/09/gotw-1-solution/

While widget w(); is clear for me, I have no idea how can the below code be a function declaration?

// same problem (gadget and doodad are types)
//
widget w( gadget(), doodad() ); // pitfall: not a variable declaration

How is this possible?

c++ c++11 gotw

share|improve this question

edited May 16 '13 at 15:21

0x499602D2

69.7k2929 gold badges123123 silver badges209209 bronze badges

asked May 16 '13 at 13:27

YankoYanko

42455 silver badges1515 bronze badges

// same problem (gadget and doodad are types)
//
widget w( gadget(), doodad() ); // pitfall: not a variable declaration

share|improve this question

edited May 16 '13 at 15:21

0x499602D2

69.7k2929 gold badges123123 silver badges209209 bronze badges

asked May 16 '13 at 13:27

YankoYanko

42455 silver badges1515 bronze badges

share|improve this question

edited May 16 '13 at 15:21

0x499602D2

69.7k2929 gold badges123123 silver badges209209 bronze badges

asked May 16 '13 at 13:27

YankoYanko

42455 silver badges1515 bronze badges

edited May 16 '13 at 15:21

0x499602D2

69.7k2929 gold badges123123 silver badges209209 bronze badges

edited May 16 '13 at 15:21

0x499602D2

69.7k2929 gold badges123123 silver badges209209 bronze badges

asked May 16 '13 at 13:27

YankoYanko

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asked May 16 '13 at 13:27

YankoYanko

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2 Answers
2

active

oldest

votes

44

In a function declaration, arguments of type array decay into pointers to the first element, arguments of type function decay into a function pointer, so the signature would be:

widget w( gadget(*)(), doodad(*)() );

That is, a function that takes as the first argument a pointer to a function taking no arguments and returning gadget, that takes as second argument a pointer to a function taking no arguments and returning a doodad and that the function itself returns a widget

There are even more interesting or confusing cases, like:

// assume 'x' is a variable defined somewhere:
widget w(gadget(x));

How could that be interpreted as a function declaration? I mean, x is a variable, right? Well, when declaring a variable you can add extra parenthesis, so gadget x; and gadget (x); both declare the same variable x. The same applies to function arguments so the code above looks like a declaration of a function that takes a first argument named x of type gadget and returns a widget...

share|improve this answer

edited May 16 '13 at 13:36

answered May 16 '13 at 13:31

David Rodríguez - dribeasDavid Rodríguez - dribeas

176k1616 gold badges240240 silver badges440440 bronze badges

• 
  
  
  

  
  47
  

  
  
  
  
  
  

  
  
  Mother of God. ಠ_ಠ
  
  – Yanko
  May 16 '13 at 13:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  +1 first time knowing that arguments of type function decay into a function pointer. thank you!
  
  – taocp
  May 16 '13 at 13:34
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  @Yanko: of course, that is why now you can use widget wgadget, doodad; => this cannot be misconstrued as a function :)
  
  – Matthieu M.
  May 16 '13 at 14:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MatthieuM. For those of you lucky enough to use C++11 :)
  
  – David Rodríguez - dribeas
  May 16 '13 at 15:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DavidRodríguez-dribeas: I am afraid I am lucky yet... at least, at work.
  
  – Matthieu M.
  May 17 '13 at 6:08
  

  
  
  
  

 | 
show 1 more comment

1

It's function that gets two functions, that returns gadget and doodad and either of them gets no arguments.

Example that compiles fine.

#include <iostream>
class widget;
class gadget;
class doodad;
gadget a()
doodad b() ;
widget w( gadget(), doodad() )

int main() 
 w(a,b);
 return 0;

http://ideone.com/YjZK9Y

share|improve this answer

edited May 16 '13 at 21:00

answered May 16 '13 at 13:33

RiaDRiaD

34.2k99 gold badges6060 silver badges103103 bronze badges

add a comment | 

44

In a function declaration, arguments of type array decay into pointers to the first element, arguments of type function decay into a function pointer, so the signature would be:

widget w( gadget(*)(), doodad(*)() );

That is, a function that takes as the first argument a pointer to a function taking no arguments and returning gadget, that takes as second argument a pointer to a function taking no arguments and returning a doodad and that the function itself returns a widget

There are even more interesting or confusing cases, like:

// assume 'x' is a variable defined somewhere:
widget w(gadget(x));

How could that be interpreted as a function declaration? I mean, x is a variable, right? Well, when declaring a variable you can add extra parenthesis, so gadget x; and gadget (x); both declare the same variable x. The same applies to function arguments so the code above looks like a declaration of a function that takes a first argument named x of type gadget and returns a widget...

share|improve this answer

edited May 16 '13 at 13:36

answered May 16 '13 at 13:31

David Rodríguez - dribeasDavid Rodríguez - dribeas

176k1616 gold badges240240 silver badges440440 bronze badges

• 
  
  
  

  
  47
  

  
  
  
  
  
  

  
  
  Mother of God. ಠ_ಠ
  
  – Yanko
  May 16 '13 at 13:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  +1 first time knowing that arguments of type function decay into a function pointer. thank you!
  
  – taocp
  May 16 '13 at 13:34
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  @Yanko: of course, that is why now you can use widget wgadget, doodad; => this cannot be misconstrued as a function :)
  
  – Matthieu M.
  May 16 '13 at 14:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MatthieuM. For those of you lucky enough to use C++11 :)
  
  – David Rodríguez - dribeas
  May 16 '13 at 15:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DavidRodríguez-dribeas: I am afraid I am lucky yet... at least, at work.
  
  – Matthieu M.
  May 17 '13 at 6:08
  

  
  
  
  

 | 
show 1 more comment

44

In a function declaration, arguments of type array decay into pointers to the first element, arguments of type function decay into a function pointer, so the signature would be:

widget w( gadget(*)(), doodad(*)() );

That is, a function that takes as the first argument a pointer to a function taking no arguments and returning gadget, that takes as second argument a pointer to a function taking no arguments and returning a doodad and that the function itself returns a widget

There are even more interesting or confusing cases, like:

// assume 'x' is a variable defined somewhere:
widget w(gadget(x));

How could that be interpreted as a function declaration? I mean, x is a variable, right? Well, when declaring a variable you can add extra parenthesis, so gadget x; and gadget (x); both declare the same variable x. The same applies to function arguments so the code above looks like a declaration of a function that takes a first argument named x of type gadget and returns a widget...

share|improve this answer

edited May 16 '13 at 13:36

answered May 16 '13 at 13:31

David Rodríguez - dribeasDavid Rodríguez - dribeas

176k1616 gold badges240240 silver badges440440 bronze badges

• 
  
  
  

  
  47
  

  
  
  
  
  
  

  
  
  Mother of God. ಠ_ಠ
  
  – Yanko
  May 16 '13 at 13:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  +1 first time knowing that arguments of type function decay into a function pointer. thank you!
  
  – taocp
  May 16 '13 at 13:34
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  @Yanko: of course, that is why now you can use widget wgadget, doodad; => this cannot be misconstrued as a function :)
  
  – Matthieu M.
  May 16 '13 at 14:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MatthieuM. For those of you lucky enough to use C++11 :)
  
  – David Rodríguez - dribeas
  May 16 '13 at 15:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @DavidRodríguez-dribeas: I am afraid I am lucky yet... at least, at work.
  
  – Matthieu M.
  May 17 '13 at 6:08
  

  
  
  
  

 | 
show 1 more comment

In a function declaration, arguments of type array decay into pointers to the first element, arguments of type function decay into a function pointer, so the signature would be:

widget w( gadget(*)(), doodad(*)() );

That is, a function that takes as the first argument a pointer to a function taking no arguments and returning gadget, that takes as second argument a pointer to a function taking no arguments and returning a doodad and that the function itself returns a widget

There are even more interesting or confusing cases, like:

// assume 'x' is a variable defined somewhere:
widget w(gadget(x));

How could that be interpreted as a function declaration? I mean, x is a variable, right? Well, when declaring a variable you can add extra parenthesis, so gadget x; and gadget (x); both declare the same variable x. The same applies to function arguments so the code above looks like a declaration of a function that takes a first argument named x of type gadget and returns a widget...

share|improve this answer

edited May 16 '13 at 13:36

answered May 16 '13 at 13:31

David Rodríguez - dribeasDavid Rodríguez - dribeas

176k1616 gold badges240240 silver badges440440 bronze badges

widget w( gadget(*)(), doodad(*)() );

// assume 'x' is a variable defined somewhere:
widget w(gadget(x));

share|improve this answer

edited May 16 '13 at 13:36

answered May 16 '13 at 13:31

David Rodríguez - dribeasDavid Rodríguez - dribeas

176k1616 gold badges240240 silver badges440440 bronze badges

answered May 16 '13 at 13:31

David Rodríguez - dribeasDavid Rodríguez - dribeas

176k1616 gold badges240240 silver badges440440 bronze badges

answered May 16 '13 at 13:31

David Rodríguez - dribeasDavid Rodríguez - dribeas

176k1616 gold badges240240 silver badges440440 bronze badges

1

It's function that gets two functions, that returns gadget and doodad and either of them gets no arguments.

Example that compiles fine.

#include <iostream>
class widget;
class gadget;
class doodad;
gadget a()
doodad b() ;
widget w( gadget(), doodad() )

int main() 
 w(a,b);
 return 0;

http://ideone.com/YjZK9Y

share|improve this answer

edited May 16 '13 at 21:00

answered May 16 '13 at 13:33

RiaDRiaD

34.2k99 gold badges6060 silver badges103103 bronze badges

add a comment | 

1

It's function that gets two functions, that returns gadget and doodad and either of them gets no arguments.

Example that compiles fine.

#include <iostream>
class widget;
class gadget;
class doodad;
gadget a()
doodad b() ;
widget w( gadget(), doodad() )

int main() 
 w(a,b);
 return 0;

http://ideone.com/YjZK9Y

share|improve this answer

edited May 16 '13 at 21:00

answered May 16 '13 at 13:33

RiaDRiaD

34.2k99 gold badges6060 silver badges103103 bronze badges

add a comment | 

It's function that gets two functions, that returns gadget and doodad and either of them gets no arguments.

Example that compiles fine.

#include <iostream>
class widget;
class gadget;
class doodad;
gadget a()
doodad b() ;
widget w( gadget(), doodad() )

int main() 
 w(a,b);
 return 0;

http://ideone.com/YjZK9Y

share|improve this answer

edited May 16 '13 at 21:00

answered May 16 '13 at 13:33

RiaDRiaD

34.2k99 gold badges6060 silver badges103103 bronze badges

#include <iostream>
class widget;
class gadget;
class doodad;
gadget a()
doodad b() ;
widget w( gadget(), doodad() )

int main() 
 w(a,b);
 return 0;

share|improve this answer

edited May 16 '13 at 21:00

answered May 16 '13 at 13:33

RiaDRiaD

34.2k99 gold badges6060 silver badges103103 bronze badges

answered May 16 '13 at 13:33

RiaDRiaD

34.2k99 gold badges6060 silver badges103103 bronze badges

answered May 16 '13 at 13:33

RiaDRiaD

34.2k99 gold badges6060 silver badges103103 bronze badges