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21

3

I have member function (method) which uses

std::enable_shared_from_this::weak_from_this() 

In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.

Is there a way to check against it at compile time?

c++ constructor c++17 shared-ptr weak-ptr

share|improve this question

edited Apr 14 at 12:27

curiousguy

4,69223046

asked Apr 8 at 14:53

KorriKorri

38629

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.
  
  – rubenvb
  Apr 8 at 14:58
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.
  
  – SergeyA
  Apr 8 at 14:59
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  @SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.
  
  – Bakuriu
  Apr 8 at 21:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bakuriu well, I did not have the energy to polish it to the full blown answer. It is possible that it is not a workable solution.
  
  – SergeyA
  Apr 9 at 14:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Simply based on the fact the static type of an object doesn't (and cannot) depend on the fact its construction is completed. Note that if that were the case, you wouldn't be able in a ctor of T to store this in a global obj register, and use the obj later, w/o a cast, as the ptr stored would have "(in construction)T*" not plain T*.
  
  – curiousguy
  Apr 14 at 12:27
  

  
  
  
  

add a comment | 

21

3

I have member function (method) which uses

std::enable_shared_from_this::weak_from_this() 

In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.

Is there a way to check against it at compile time?

c++ constructor c++17 shared-ptr weak-ptr

share|improve this question

edited Apr 14 at 12:27

curiousguy

4,69223046

asked Apr 8 at 14:53

KorriKorri

38629

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.
  
  – rubenvb
  Apr 8 at 14:58
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.
  
  – SergeyA
  Apr 8 at 14:59
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  @SergeyA Why didn't you post that as an answer? All other people here seem to conclude that it's not possible so either your comment is wrong and misleading or they are wrong and you should show how it can be achieved.
  
  – Bakuriu
  Apr 8 at 21:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bakuriu well, I did not have the energy to polish it to the full blown answer. It is possible that it is not a workable solution.
  
  – SergeyA
  Apr 9 at 14:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Simply based on the fact the static type of an object doesn't (and cannot) depend on the fact its construction is completed. Note that if that were the case, you wouldn't be able in a ctor of T to store this in a global obj register, and use the obj later, w/o a cast, as the ptr stored would have "(in construction)T*" not plain T*.
  
  – curiousguy
  Apr 14 at 12:27
  

  
  
  
  

add a comment | 

I have member function (method) which uses

std::enable_shared_from_this::weak_from_this() 

In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.

Is there a way to check against it at compile time?

c++ constructor c++17 shared-ptr weak-ptr

share|improve this question

edited Apr 14 at 12:27

curiousguy

4,69223046

asked Apr 8 at 14:53

KorriKorri

38629

std::enable_shared_from_this::weak_from_this() 

share|improve this question

edited Apr 14 at 12:27

curiousguy

4,69223046

asked Apr 8 at 14:53

KorriKorri

38629

share|improve this question

edited Apr 14 at 12:27

curiousguy

4,69223046

asked Apr 8 at 14:53

KorriKorri

38629

edited Apr 14 at 12:27

curiousguy

4,69223046

edited Apr 14 at 12:27

curiousguy

4,69223046

asked Apr 8 at 14:53

KorriKorri

38629

asked Apr 8 at 14:53

KorriKorri

38629

3 Answers
3

active

oldest

votes

15

I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.

share|improve this answer

answered Apr 8 at 14:59

AngewAngew

135k11261357

add a comment | 

5

No there is no way. Consider:

void call_me(struct widget*);

struct widget : std::enable_shared_from_this<widget> 
 widget() 
 call_me(this);
 

 void display() 
 shared_from_this();
 
;

// later:

void call_me(widget* w) 
 w->display(); // crash

The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.

share|improve this answer

answered Apr 8 at 15:37

Guillaume RacicotGuillaume Racicot

16.7k53873

add a comment | 

4

Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:

class A 

 // ... whatever ...
public:
 A() 
 // do construction work
 constructed = true;
 

 foo() 
 if (not constructed) 
 throw std::logic_error("Cannot call foo() during construction"); 
 
 // the rest of foo
 

protected:
 bool constructed false ;

You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.

An alternative to throwing could be assert()'ing.

share|improve this answer

edited Apr 9 at 7:29

answered Apr 8 at 16:58

einpoklumeinpoklum

38.2k28135266

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.
  
  – Korri
  Apr 8 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.
  
  – Jesper Juhl
  Apr 8 at 18:38
  
  

  
  
  
  

add a comment | 

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15

I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.

share|improve this answer

answered Apr 8 at 14:59

AngewAngew

135k11261357

add a comment | 

15

I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.

share|improve this answer

answered Apr 8 at 14:59

AngewAngew

135k11261357

add a comment | 

I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.

share|improve this answer

answered Apr 8 at 14:59

AngewAngew

135k11261357

share|improve this answer

answered Apr 8 at 14:59

AngewAngew

135k11261357

answered Apr 8 at 14:59

AngewAngew

135k11261357

answered Apr 8 at 14:59

AngewAngew

135k11261357

5

No there is no way. Consider:

void call_me(struct widget*);

struct widget : std::enable_shared_from_this<widget> 
 widget() 
 call_me(this);
 

 void display() 
 shared_from_this();
 
;

// later:

void call_me(widget* w) 
 w->display(); // crash

The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.

share|improve this answer

answered Apr 8 at 15:37

Guillaume RacicotGuillaume Racicot

16.7k53873

add a comment | 

5

No there is no way. Consider:

void call_me(struct widget*);

struct widget : std::enable_shared_from_this<widget> 
 widget() 
 call_me(this);
 

 void display() 
 shared_from_this();
 
;

// later:

void call_me(widget* w) 
 w->display(); // crash

The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.

share|improve this answer

answered Apr 8 at 15:37

Guillaume RacicotGuillaume Racicot

16.7k53873

add a comment | 

No there is no way. Consider:

void call_me(struct widget*);

struct widget : std::enable_shared_from_this<widget> 
 widget() 
 call_me(this);
 

 void display() 
 shared_from_this();
 
;

// later:

void call_me(widget* w) 
 w->display(); // crash

The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.

share|improve this answer

answered Apr 8 at 15:37

Guillaume RacicotGuillaume Racicot

16.7k53873

void call_me(struct widget*);

struct widget : std::enable_shared_from_this<widget> 
 widget() 
 call_me(this);
 

 void display() 
 shared_from_this();
 
;

// later:

void call_me(widget* w) 
 w->display(); // crash

share|improve this answer

answered Apr 8 at 15:37

Guillaume RacicotGuillaume Racicot

16.7k53873

answered Apr 8 at 15:37

Guillaume RacicotGuillaume Racicot

16.7k53873

answered Apr 8 at 15:37

Guillaume RacicotGuillaume Racicot

16.7k53873

4

Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:

class A 

 // ... whatever ...
public:
 A() 
 // do construction work
 constructed = true;
 

 foo() 
 if (not constructed) 
 throw std::logic_error("Cannot call foo() during construction"); 
 
 // the rest of foo
 

protected:
 bool constructed false ;

You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.

An alternative to throwing could be assert()'ing.

share|improve this answer

edited Apr 9 at 7:29

answered Apr 8 at 16:58

einpoklumeinpoklum

38.2k28135266

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.
  
  – Korri
  Apr 8 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.
  
  – Jesper Juhl
  Apr 8 at 18:38
  
  

  
  
  
  

add a comment | 

4

Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:

class A 

 // ... whatever ...
public:
 A() 
 // do construction work
 constructed = true;
 

 foo() 
 if (not constructed) 
 throw std::logic_error("Cannot call foo() during construction"); 
 
 // the rest of foo
 

protected:
 bool constructed false ;

You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.

An alternative to throwing could be assert()'ing.

share|improve this answer

edited Apr 9 at 7:29

answered Apr 8 at 16:58

einpoklumeinpoklum

38.2k28135266

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.
  
  – Korri
  Apr 8 at 17:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.
  
  – Jesper Juhl
  Apr 8 at 18:38
  
  

  
  
  
  

add a comment | 

Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:

class A 

 // ... whatever ...
public:
 A() 
 // do construction work
 constructed = true;
 

 foo() 
 if (not constructed) 
 throw std::logic_error("Cannot call foo() during construction"); 
 
 // the rest of foo
 

protected:
 bool constructed false ;

You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.

An alternative to throwing could be assert()'ing.

share|improve this answer

edited Apr 9 at 7:29

answered Apr 8 at 16:58

einpoklumeinpoklum

38.2k28135266

class A 

 // ... whatever ...
public:
 A() 
 // do construction work
 constructed = true;
 

 foo() 
 if (not constructed) 
 throw std::logic_error("Cannot call foo() during construction"); 
 
 // the rest of foo
 

protected:
 bool constructed false ;

share|improve this answer

edited Apr 9 at 7:29

answered Apr 8 at 16:58

einpoklumeinpoklum

38.2k28135266

answered Apr 8 at 16:58

einpoklumeinpoklum

38.2k28135266

answered Apr 8 at 16:58

einpoklumeinpoklum

38.2k28135266